Let $x \ y^2=u$ and $z=2x=v$ hence $f(u,v)=0$

$\therefore \dfrac {\partial f}{\partial x} = \dfrac {\partial f}{\partial u}\cdot \dfrac {\partial u}{\partial x}+ \dfrac {\partial f}{\partial v}\cdot \dfrac {\partial v}{\partial x}=0 \ \cdots \ \cdots (1)$

$ \therefore \dfrac {\partial f}{\partial y} = \dfrac {\partial f}{\partial u}\cdot \dfrac {\partial u}{\partial x} + \dfrac {\partial f}{\partial v} \cdot \dfrac {\partial v}{\partial y}=0 \cdots \ \cdots (2)$

But, $\dfrac {\partial u}{\partial x}=y^2 $ and $\dfrac {\partial u}{\partial y}=2xy$

And, $ \dfrac{\partial v}{\partial x} = \dfrac {\partial z}{\partial x} -2 $ and $ \dfrac {\partial v}{\partial y} = \dfrac {\partial z}{\partial y}$

From (1) and (2), we have

$\dfrac {\partial f}{\partial u}\cdot y^2 + \dfrac {\partial f}{\partial v} \left ( \dfrac {\partial z}{\partial x}-2 \right )=0 \ \cdots \ \cdots (3)$

And, $\dfrac {\partial f}{\partial u}\cdot 2xy+\dfrac {\partial f}{\partial v}\left ( \dfrac {\partial z}{\partial y} \right ) = 0 \ \cdots \ \cdots (4)$

From (3) and (4), we get

$ \dfrac {\left ( \frac {\partial f}{\partial u} \right )}{\left ( \frac {\partial f}{\partial v} \right )} = - \dfrac {\left ( \frac {\partial z}{\partial x}-2 \right )}{y^2} \text{ and } \dfrac {\left ( \frac {\partial f}{\partial u} \right )}{\left ( \frac {\partial f}{\partial v} \right )}= - \dfrac {\left ( \frac {\partial z}{\partial y} \right )}{2xy}$

∴ Equating the two,

$\dfrac {\left ( \dfrac {\partial z}{\partial x}-2 \right )}{y^2} = \dfrac {\left ( \dfrac {\partial z}{\partial y} \right )}{2xy}$

$\therefore 2x \dfrac {\partial z}{\partial x} - 4x = y \dfrac {\partial z}{\partial y}$

$\therefore 2x \dfrac {\partial z}{\partial x} - y \dfrac {\partial z}{\partial y}= 4x$