$x u_x\+y u_y\+z u_z$

$\=x \left(x^3\left(F_r r_x\+F_s s_x\right)\+3 x^{2}F\right)\+y x^3 \left(F_r r_y\+F_s s_y\right)\+z x^3 \left(F_r r_z\+F_s s_z\right)$

$\=x \left(x^3\left(F_r \left(-\frac{y}{x^2}\right)\+F_s \left(-\frac{z}{x^2}\right)\right)\+3 x^{2}F\right)\+y x^3 \left(F_r \left(\frac{1}{x}\right)\+F_s \left(0\right)\right)\+z x^3 \left(F_r \left(0\right)\+F_s \left(\frac{1}{x}\right)\right)$

$\=\left(-x^{2}y F_r-x^{2}z F_s\+3 x^{3}F\right)\+\left(x^{2}y F_r\+0\right)\+\left(0\+x^{2}z F_s\right)$

$\=\left(-x^{2}y\+x^{2}y\right)F_r\+\left(-x^{2}z\+x^{2}z\right)F_s\+3 u$

$\=0\cdot F_r\+0\cdot F_s\+3 u$

$\=0\+0\+3 u$

$\=3 u$

Define $r\,s$, and $u$ 

$r\=y\/x$$\stackrel{\text{assign}}{\to }$

$s\=z\/x$$\stackrel{\text{assign}}{\to }$

$u\=x^3 F\left(r\,s\right)$$\stackrel{\text{assign}}{\to }$

Compute $x u_x\+y u_y\+z u_z$ and simplify the result to $3 x^{3}F\=3 u$

$x \frac{\partial }{\partial x} u\+y \frac{\partial }{\partial y} u\+z \frac{\partial }{\partial z} u$ = $x\left(3x^{2}F\left(\frac{y}{x}\,\frac{z}{x}\right)\+x^3\left(-\frac{{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)y}{x^2}-\frac{{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)z}{x^2}\right)\right)\+y{x}^2{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)\+z{x}^2{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)$$\stackrel{\text{simplify}}{\=}$$3x^{3}F\left(\frac{y}{x}\,\frac{z}{x}\right)$

Separately obtain and simplify the partial derivatives $u_x\,u_y$, and $u_z$ 

$\frac{\partial }{\partial x} u$ = $3x^{2}F\left(\frac{y}{x}\,\frac{z}{x}\right)\+x^3\left(-\frac{{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)y}{x^2}-\frac{{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)z}{x^2}\right)$$\stackrel{\text{expand}}{\=}$$3x^{2}F\left(\frac{y}{x}\,\frac{z}{x}\right)-x{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)y-x{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)z$$$

$\frac{\partial }{\partial y} u$ = $x^2{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)$$$

$\frac{\partial }{\partial z} u$ = $x^2{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)$$$

Obtain and simplify $r_x\,r_y\,r_z\,s_x\,s_y$, and $s_z$ 

$\frac{\partial }{\partial x} r$ = $-\frac{y}{x^2}$

$\frac{\partial }{\partial x} s$ = $-\frac{z}{x^2}$

$\frac{\partial }{\partial y} r$ = $\frac{1}{x}$

$\frac{\partial }{\partial y} s$ = $0$$$

$\frac{\partial }{\partial z} r$ = $0$$$

$\frac{\partial }{\partial z} s$ = $\frac{1}{x}$

Initialize

$r≔y\/x\:$

$s≔z\/x\:$

$u≔x^{3}F\left(r\,s\right)\:$

Apply the diff and simplify commands to evaluate the expression $x u_x\+y u_y\+z u_z$ 

$\mathrm{simplify}\left(x \mathrm{diff}\left(u\,x\right)\+y \mathrm{diff}\left(u\,y\right)\+z \mathrm{diff}\left(u\,z\right)\right)$ = $3x^{3}F\left(\frac{y}{x}\,\frac{z}{x}\right)$$$

Separately obtain and simplify $u_x\,u_y$, and $u_z$ 

$\mathrm{expand}\left(\mathrm{diff}\left(u\,x\right)\right)$ = $3x^{2}F\left(\frac{y}{x}\,\frac{z}{x}\right)-x{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)y-x{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)z$$$

$\mathrm{diff}\left(u\,y\right)$ = $x^2{\mathrm{D}}_1\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)$$$

$\mathrm{diff}\left(u\,z\right)$ = $x^2{\mathrm{D}}_2\left(F\right)\left(\frac{y}{x}\,\frac{z}{x}\right)$$$

Obtain and simplify the derivatives $r_x\,r_y\,r_z$ and $s_x\,s_y\,s_z$ 

$\mathrm{diff}\left(r\,x\right)$ = $-\frac{y}{x^2}$$$

$\mathrm{diff}\left(s\,x\right)$ = $-\frac{z}{x^2}$$$

$\mathrm{diff}\left(r\,y\right)$ = $\frac{1}{x}$$$

$\mathrm{diff}\left(s\,y\right)$ = $0$$$

$\mathrm{diff}\left(r\,z\right)$ = $0$$$

$\mathrm{diff}\left(s\,z\right)$ = $\frac{1}{x}$$$