If $F(xy^2,z-2x)=0$ then prove $x\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y }.\frac{y}{2}=2x$

Question

If $F(xy^2,z-2x)=0$ then prove $$x\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y }.\frac{y}{2}=2x$$ how can we get this expression ? and I don't understand what use of $F?$ I can substitute $xy^2$ and $z-2x$ and then try differentiation but that doesn't seem viable here since it will add 2 more variable which maybe unnecessary?

Answer 1

The relationship $F(xy^2,z-2x)=0$ implicitly defines $z$ as a function of $x$ and $y$. Differentiating gives: \begin{align*} 0&=y^2F_1+\left(\frac{\partial z}{\partial x}-2\right)F_2,\\ 0&=2xyF_1+\frac{\partial z}{\partial y}F_2 \end{align*} where $F_1$ and $F_2$ are, respectively, the partial derivatives of $F$ with respect to its first input and second input. When we multiply the first equality above by $2x$ and the second by $y$, the equality of the coefficients of $F_2$ implies $$ 2x\left(\frac{\partial z}{\partial x}-2\right)=y\frac{\partial z}{\partial y}\iff x\frac{\partial z}{\partial x}-\frac{y}{2}\frac{\partial z}{\partial y}=2x. $$