Answer
\[y^{'}=\frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}\]
Work Step by Step
$\sqrt{xy}=1+x^2y$
$\sqrt{x}\sqrt{y}=1+x^2y$ ___(1)
Differentiating (1) implicitly with respect to $x$
$(\sqrt{x})^{'}\sqrt{y}+\sqrt{x}(\sqrt{y})^{'}=(x^2)^{'}y+x^2(y)^{'}$
$\large\frac{1}{2\sqrt{x}}$ $\sqrt{y}$+$\large\frac{\sqrt{x}}{2\sqrt{y}}$ $y^{'}$ $=2xy+x^2y^{'}$
$\frac{\sqrt{y}}{2\sqrt{x}}-2xy=y^{'}\left[x^2-\frac{\sqrt{x}}{2\sqrt{y}}\right]$
$\Large\frac{\sqrt{y}-4x\sqrt{x}y}{2\sqrt{x}}$ $=y^{'}\Large\left[\frac{2x^2\sqrt{y}-\sqrt{x}}{2\sqrt{y}}\right]$
$y^{'}=\Large\frac{\sqrt{y}}{\sqrt{x}}.\left[\frac{\sqrt{y}-4x\sqrt{x}y}{2x^2\sqrt{y}-\sqrt{x}}\right]$
$y^{'}=\Large\frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$
Hence, $y^{'}=\Large\frac{4xy\sqrt{xy}-y}{x-2x^2\sqrt{xy}}$.
