#### Answer

$s_u=\dfrac{1+2y}{1-2y}$

#### Work Step by Step

Differentiate $x$ and $y$ with respect to $u$.
$1=2x \times x_u-2yy_u .....(a)$ and $0=2x \times x_u -y_u ....(b)$
Now, $[0=2x \times x_u -y_u ....(b)] \times 2y$ and subtract from equation $a$.
So, $x_u= \dfrac{1}{2x-4xy}$
Subtract equation (b) from equation $a$.
We get $y_u= \dfrac{1}{1-2y}$
Now, differentiate $s$ with respect to $u$.
$s_u=2x \times x_u-2y \times y_u =2x \times ( \dfrac{1}{2x-4xy}) -2y \times (\dfrac{1}{2y})$
So, $s_u=\dfrac{1+2y}{1-2y}$