$x{e}^{xy}=y+{\sin }^{2}x,$ 

so at $x=0$, 

$0·e^{0·y}=y+{\sin }^2\left(0\right)\Rightarrow y=0$ ...(1)

Now, differentiating both sides

i.e. $\frac{d\left(x{e}^{xy}\right)}{dx}=\frac{dy}{dx}+\frac{d\left({\sin }^{2}x\right)}{dx}$

$\Rightarrow x{e}^{xy}\left(x\frac{dy}{dx}+y\right)+e^{xy}=\frac{dy}{dx}+2\sin x\cos x$ (Using chain rule i.e. ${\left[f\left(g\left(x\right)\right)\right]}^{\text{'}}=f^{\text{'}}\left(g\left(x\right)\right)·g^{\text{'}}\left(x\right)$ and product rule i.e. ${\left(u·v\right)}^{\text{'}}=u{v}^{\text{'}}+u^{\text{'}}v$)

Putting $x=0$,

$\Rightarrow 0·e^0\left(0·{\left(\frac{dy}{dx}\right)}_{x=0}+0\right)+e^0={\left(\frac{dy}{dx}\right)}_{x=0}+2\sin \left(0\right)\cos \left(0\right)$

$\Rightarrow 0+e^0={\left(\frac{dy}{dx}\right)}_{x=0}+0$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0}=1$