Answer
$$135e^{12}$$
Work Step by Step
Given $$h(u, v)=u e^{v}, \ \ \ u=q^{3},\ \ \ \ \ v=q r^{2}$$
Since at $(q,r)=(3,2)$; $(u,v)= ( 27,12)$
$$
\frac{\partial h}{\partial \:u}= e^{v},\ \ \ \ \ \frac{\partial h}{\partial \:v}=u e^{v} \\ \frac{\partial u}{\partial q}=3q^2,\ \ \ \ \ \ \ \ \ \ \ \frac{\partial v}{\partial q}=r^2
$$
Then
\begin{align*}
\frac{\partial h }{ \partial u}&=\frac{\partial h }{ \partial u}\frac{\partial u }{ \partial q}+\frac{\partial h }{ \partial v}\frac{\partial v }{ \partial q}\\
&= 3q^2e^v+ur^2e^v \\
&=3q^2e^{qr^2}+q^3r^2e^{qr^2}
\end{align*}
Hence
\begin{align*}
\frac{\partial h }{ \partial q}\bigg|_{(3,2)}&= 3(3)^2e^{(3)(2)^2}+(3)^3(2)^2e^{(3)(2)^2}\\
&=135e^{12}
\end{align*}
