$\underset{x\to 0}{lim}f\left(x\,m x\right)$

$\=\underset{x\to 0}{lim}\frac{3m^3{x}^4\+2m^2{x}^4\+x^4}{{\left(m^2{x}^2\+x^2\right)}^2}$

$\=\underset{x\to 0}{lim}\frac{3m^3\+2m^2\+1}{{\left(m^2\+1\right)}^2}$

$\=\frac{3m^3\+2m^2\+1}{{\left(m^2\+1\right)}^2}$

Define the function $f\left(x\,y\right)$ 

$f\left(x\,y\right)\=\frac{x^4\+2 x^2{y}^2\+3 x y^3}{{\left(x^2\+y^2\right)}^2}$$\stackrel{\text{assign as function}}{\to }$$f$

Evaluate $\underset{x\to 0}{lim}f\left(x\,m x\right)$ 

$\underset{x\to 0}{lim}f\left(x\,m x\right)$ = $\frac{3m^3+2m^2+1}{{\left(m^2+1\right)}^2}$

$f\left(x\,y\right)$ = $\frac{x^4+2x^2{y}^2+3x{y}^3}{{\left(x^2+y^2\right)}^2}$$\stackrel{\text{bivariate limit}}{\to }$$\frac{71}{160}-\frac{31\sqrt{31}}{160}..\frac{71}{160}+\frac{31\sqrt{31}}{160}$

$f≔\left(x\,y_{}\right)\to \frac{x^4\+2 x^2{y}^2\+3 x y^3}{{\left(x^2\+y^2\right)}^2}\:$

$\mathrm{limit}\left(f\left(x\,m x\right)\,x\=0\right)$ = $\frac{3m^3+2m^2+1}{{\left(m^2+1\right)}^2}$

For corroboration, apply Maple's bivariate limit command

$\mathrm{limit}\left(f\left(x\,y\right)\,\left\{x\=0\,y\=0\right\}\right)$ = $\frac{71}{160}-\frac{31\sqrt{31}}{160}..\frac{71}{160}+\frac{31\sqrt{31}}{160}$