$$$f_x\=g_z z_x\=0$ ⇒ $z_x\= \frac{f_x}{g_z}$$$

and

$f_y\=g_y\+g_z z_y\=0$ ⇒ $z_x\= \frac{f_y-g_y}{g_z}$$$

Solution via the Context Panel

Tools≻Tasks≻Browse:

Calculus - Differential≻Derivatives≻Implicit Differentiation≻$f\left(x\,y\,z\right)\=g\left(x\,y\,z\right)$ 

Table 4.3.21(a)   Implicit differentiation task template

$f\left(x\,y\right)\=g\left(y\,z\right)$

$f\left(x\,y\right)\=g\left(y\,z\right)$

$\stackrel{\text{evaluate at point}}{\to }$

$f\left(x\,y\right)\=g\left(y\,z\left(x\,y\right)\right)$

$\stackrel{\text{differentiate w.r.t. x}}{\to }$

$\frac{\partial }{\partial x}f\left(x\,y\right)\={\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}z\left(x\,y\right)\right)$

$\stackrel{\text{isolate for diff(z(x,y),x)}}{\to }$

$\frac{\partial }{\partial x}z\left(x\,y\right)\=\frac{\frac{\partial }{\partial x}f\left(x\,y\right)}{{\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)}$

Table 4.3.21(b)   Context Panel implementation of implicit derivative via first principles

Immediate results via the implicitdiff command

$\mathrm{implicitdiff}\left(f\left(x\,y\right)\=g\left(y\,z\right)\,z\left(x\,y\right)\,x\right)$ = $\frac{{\mathrm{D}}_1\left(f\right)\left(x\,y\right)}{{\mathrm{D}}_2\left(g\right)\left(y\,z\right)}$$$

$\mathrm{implicitdiff}\left(f\left(x\,y\right)\=g\left(y\,z\right)\,z\left(x\,y\right)\,y\right)$ = $-\frac{-{\mathrm{D}}_2\left(f\right)\left(x\,y\right)\+{\mathrm{D}}_1\left(g\right)\left(y\,z\right)}{{\mathrm{D}}_2\left(g\right)\left(y\,z\right)}$$$

Solution from first principles

$\mathrm{temp}≔\mathrm{diff}\left(f\left(x\,y\right)\=g\left(y\,z\left(x\,y\right)\right)\,x\right)$

$\frac{\partial }{\partial x}f\left(x\,y\right)\={\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)\left(\frac{\partial }{\partial x}z\left(x\,y\right)\right)$

$\mathrm{isolate}\left(\mathrm{temp}\,\mathrm{diff}\left(z\left(x\,y\right)\,x\right)\right)$

$\frac{\partial }{\partial x}z\left(x\,y\right)\=\frac{\frac{\partial }{\partial x}f\left(x\,y\right)}{{\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)}$

$\mathrm{temp}≔\mathrm{diff}\left(f\left(x\,y\right)\=g\left(y\,z\left(x\,y\right)\right)\,y\right)$

$\frac{\partial }{\partial y}f\left(x\,y\right)\={\mathrm{D}}_1\left(g\right)\left(y\,z\left(x\,y\right)\right)\+{\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}z\left(x\,y\right)\right)$

$\mathrm{isolate}\left(\mathrm{temp}\,\mathrm{diff}\left(z\left(x\,y\right)\,y\right)\right)$

$\frac{\partial }{\partial y}z\left(x\,y\right)\=-\frac{-\left(\frac{\partial }{\partial y}f\left(x\,y\right)\right)\+{\mathrm{D}}_1\left(g\right)\left(y\,z\left(x\,y\right)\right)}{{\mathrm{D}}_2\left(g\right)\left(y\,z\left(x\,y\right)\right)}$