See Differentialgleichungen, by E. Kamke, p. 30. The technique consists mainly of looking for a parametric solution. Consider, for instance, the case y=G(x,diff(y,x)).

ode := y=G(x,diff(y(x),x));

$\mathrm{ode}≔y=G\left(x\,\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)$

Choosing the parametrization

para := diff(y(x),x) = t;

$\mathrm{para}≔\frac{\ⅆ}{\ⅆx}y\left(x\right)=t$

ode1 := subs(para,x=x(t),y=y(t), ode);

$\mathrm{ode1}≔y\left(t\right)=G\left(x\left(t\right)\,t\right)$

From the equations above and using the chain rule $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}}$ $\frac{\mathrm{dt}}{\mathrm{dx}}$, it is possible to obtain another ODE for $x\left(t\right)$ as follows:

ode_draft := diff(x(t),t) = diff(rhs(ode1),t)/t:

ode2 := diff(x(t),t)=solve(ode_draft,diff(x(t),t));

$\mathrm{ode2}≔\frac{\ⅆ}{\ⅆt}x\left(t\right)=-\frac{{\mathrm{D}}_2\left(G\right)\left(x\left(t\right)\,t\right)}{{\mathrm{D}}_1\left(G\right)\left(x\left(t\right)\,t\right)-t}$

You should therefore solve $\mathrm{ode2}$ for $x\left(t\right)$ and determine $y\left(t\right)$ by introducing the resulting $x\left(t\right)$ in $\mathrm{ode1}$. Note that, when G does not depend on $x\left(t\right),\mathrm{ode2}$ is a quadrature. ODEs matching the pattern $x=G\left(y\,\frac{\ⅆy}{\ⅆx}\right)$ are solved using the same ideas, and ODEs matching the patterns $0=G\left(x\,\frac{\ⅆy}{\ⅆx}\right),0=G\left(y\,\frac{\ⅆy}{\ⅆx}\right),x=G\left(\frac{\ⅆy}{\ⅆx}\right)$, or $y=G\left(\frac{\ⅆy}{\ⅆx}\right)$ are just particular cases.

Although any ODE can be attempted using the scheme outlined above, generally speaking, there are four cases which can be better dealt with by looking for a parametric solution; they are:

$y=G\left(x\,\frac{\ⅆy}{\ⅆx}\right)$

$x=G\left(y\,\frac{\ⅆy}{\ⅆx}\right)$

$y=G\left(\frac{\ⅆy}{\ⅆx}\right)$   (particular case)

$x=G\left(\frac{\ⅆy}{\ⅆx}\right)$   (particular case)

Parametric solutions are available by giving the optional argument 'parametric' to dsolve. By default, when the ODE is of high degree in $\frac{\mathrm{dy}}{\mathrm{dx}}$, dsolve tries the parametric scheme, along with a set of related methods for this type of ODE. However, this scheme may also be of help in some cases in which $\frac{\mathrm{dy}}{\mathrm{dx}}$ can be isolated.

$\mathrm{with}\left(\mathrm{DEtools}\,\mathrm{odeadvisor}\right)$

$\left[\mathrm{odeadvisor}\right]$

$\mathrm{ode}≔x^{n-1}{\mathrm{diff}\left(y\left(x\right)\,x\right)}^n-nx\mathrm{diff}\left(y\left(x\right)\,x\right)+y\left(x\right)$

$\mathrm{ode}≔x^{n-1}{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^n-nx\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)+y\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode}\right)$

$\left[\mathrm{y=\_G(x,y\text{'})}\right]$

$\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{ode}\right)$

$\mathrm{ans}≔y\left(x\right)=-{\left(\frac{\mathrm{c\_\_1}{\left(\frac{x}{\mathrm{c\_\_1}}\right)}^{\frac{1}{n}}}{x}\right)}^n{x}^{n-1}+n\mathrm{c\_\_1}{\left(\frac{x}{\mathrm{c\_\_1}}\right)}^{\frac{1}{n}}$

$\mathrm{odetest}\left(\mathrm{ans}\,\mathrm{ode}\right)$

$0$

$\mathrm{ode}≔16{y\left(x\right)}^2{\mathrm{diff}\left(y\left(x\right)\,x\right)}^3+2x\mathrm{diff}\left(y\left(x\right)\,x\right)-y\left(x\right)$

$\mathrm{ode}≔16{y\left(x\right)}^2{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^3+2x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-y\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode}\right)$

$\left[\left[\mathrm{\_1st\_order}\,\mathrm{\_with\_linear\_symmetries}\right]\right]$

$\mathrm{dsolve}\left(\mathrm{ode}\right)$

$y\left(x\right)=-\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{3}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(-x^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{3},y\left(x\right)=\frac{2^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{3}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(-x^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{3},y\left(x\right)=-\frac{\mathrm{I}}{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{3}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(-x^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.},y\left(x\right)=\frac{\mathrm{I}}{3}{2}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{3}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\left(-x^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.},y\left(x\right)=\sqrt{16{\mathrm{c\_\_1}}^3+2\mathrm{c\_\_1}x},y\left(x\right)=-\sqrt{16{\mathrm{c\_\_1}}^3+2\mathrm{c\_\_1}x}$

$\mathrm{ode}≔{\mathrm{diff}\left(y\left(x\right)\,x\right)}^2\sin \left(\mathrm{diff}\left(y\left(x\right)\,x\right)\right)-y\left(x\right)$

$\mathrm{ode}≔{\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}^2\sin \left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)-y\left(x\right)$

$\mathrm{odeadvisor}\left(\mathrm{ode}\right)$

$\left[\mathrm{\_quadrature}\right]$

$\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{ode}\right)$

$\mathrm{ans}≔y\left(x\right)=0,x-\left({\int }_{\phantom{\mathrm{`\; `}}}^{y\left(x\right)}\frac{1}{\mathrm{RootOf}\left(\sin \left(\mathrm{\_Z}\right){\mathrm{\_Z}}^2-\mathrm{\_a}\right)}\ⅆ\mathrm{\_a}\right)-\mathrm{c\_\_1}=0$

$\mathrm{map}\left(\mathrm{odetest}\,\left[\mathrm{ans}\right]\,\mathrm{ode}\right)$

$\left[0\,0\right]$