Answer
$u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$ and $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$
Hence, $u_{xy}=u_{yx}$
Work Step by Step
Consider the function $u=e^{xy}siny$
Need to prove the conclusion of Clairaut’s Theorem holds, that is, $u_{xy}=u_{yx}$
In order to find this differentiate the function with respect to $x$ keeping $y$ constant.
$u_{x}=ye^{xy}siny$
Differentiate the function with respect to $x$ keeping $y$ constant.
$u_{y}=xe^{xy}siny+ye^{xy}cosy$
Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant.
$u_{xy}=\frac{∂}{∂y}[ye^{xy}siny]$ (apply product rule)
$=e^{xy}siny\frac{∂y}{∂y}+ysiny\frac{∂e^{xy}}{∂y}+ye^{xy}\frac{∂(siny)}{∂y}$
$=e^{xy}siny+ysiny(xe^{xy})+ye^{xy}cosy$
Thus, $u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$
Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant.
$u_{yx}=\frac{∂}{∂x}[xe^{xy}siny+ye^{xy}cosy]$
$=\frac{∂}{∂x}[xe^{xy}siny]+\frac{∂}{∂x}[ye^{xy}cosy]$
$=e^{xy}siny\frac{∂}{∂x}[x]+x\times\frac{∂}{∂x}[e^{xy}siny]+ye^{xy}cosy$
$=e^{xy}siny+xysiny(e^{xy})+ye^{xy}cosy$
Thus, $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$
Therefore, $u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$ and $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$
Hence, $u_{xy}=u_{yx}$
