As u is the function of x and y

$du=\dfrac {\partial u}{\partial x}dx + \dfrac {\partial u} {\partial y}dy $

$\dfrac {du}{dx} = \dfrac {\partial u}{\partial x} + \dfrac {\partial u}{\partial y}\dfrac {d y}{d x} \ \cdots (A)$

$Differentiating \ a^2x^2+b^2y^2=c^2 \ implicitly,$

$a^2(2x)+b^2 (2y)\dfrac {dy}{dx}=0 $

$ \dfrac {dy}{dx}= \dfrac {a^2 x}{b^2 y} \ \cdots (B)$

$Differentiating \ u=\sin (x^2+y^2) \ partially \ w.r.t \ x,$

$\dfrac {\partial u}{\partial x} =\cos (x^2+y^2)\times \dfrac {\partial (x^2+y^2)}{\partial x} =2x \cos (x^2 +y^2) \ \cdots (C)$

$Differentiating \ u=\sin (x^2+y^2) \ partially \ w.r.t \ y,$

$\dfrac {\partial u}{\partial y} =\cos (x^2+y^2)\times \dfrac {\partial (x^2+y^2)}{\partial x} =2x \cos (x^2 +y^2) \ \cdots (D)$

From (A), (B), (C) and (D),

$\dfrac {du}{dx}= 2x \cos (x^2 +y^2) + 2y \cos (x^2 +y^2) \times \left ( - \dfrac {a^2 x}{b^2 y} \right )$

$=2x\cos (x^2+ y^2)- \dfrac {2a^2x}{b^2} \cos (x^2 +y^2) $

$= 2x \cos (x^2 + y^2 ) \left ( 1-\dfrac {a^2}{b^2} \right )$

$\therefore \ \dfrac {du}{dx} = \dfrac {2x (b^2 -a^2)}{b^2} \cos (x^2 +y^2)$