I see such a question:
Assume that $z$ is a non-constant function of $x$ and $y$ and $\phi$ is given implicitly by $\phi(2x-z^2,y-\frac{1}{3}z^3)=0$. Show that $\frac{\partial z}{\partial x} + z\frac{\partial z}{\partial y} = \frac{1}{z}$.
I achieved to show the equation by substituting $u= 2x-z^2$ and $v=y-\frac{1}{3}z^3$ and taking the partial derivatives w.r.t $x$ and $y$;
$\phi_u(2-2\cdot z \cdot z_x) + \phi_v(-z^2z_x) = 0$ etc.
However, the solution key has a different solution that confused me a bit.
The solution is given as: $u= 2x-z^2$ and $v=y-\frac{1}{3}z^3$
$F_x = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial x}= \phi_u \cdot 2 + \phi_v \cdot 0$
$F_y = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial y}= \phi_u \cdot 0 + \phi_v \cdot 1$
$F_z = \frac{\partial \phi}{\partial u} \frac{\partial u}{\partial z} + \frac{\partial \phi}{\partial v}\frac{\partial v}{\partial z}= \phi_u \cdot (-2z) + \phi_v \cdot (-z^2)$
And then used the formula $\frac{\partial z}{\partial x} = - \frac{F_x}{F_z} $ and $\frac{\partial z}{\partial y} = - \frac{F_y}{F_z}$.
So my question is, in spite of the fact that $z$ is a function of $x$ and $y$, why did we take $\frac{\partial u}{\partial x} = 2$ instead of $2-2zz_x$, why did we take $\frac{\partial u}{\partial y} = 1$ instead of $-z^2z_y$?
You did the problem by using implicit differentiation (treating $z=z(x,y)$ when you differentiated). They used the general formula for implicit differentiation coming out of the implicit function theorem.
Note that their $F(x,y,z) = \phi(2x-z^2,y-\frac13z^3)$. You differentiated $F(x,y,z(x,y))$ directly. Your formula is identical to theirs.