Answer
$2ve^{2u/v}$ and $-2ue^{2u/v}+2ve^{2u/v}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $g$ with respect to $u$, by keeping $v$ as a constant, and vice versa:
$g_u=\dfrac{2}{v} \times v^2 \space e^{2u/v}=2ve^{2u/v}$
and
$g_v=v^2e^{2(u/v)}\times [-2 \space u/v^2]+2ve^{2u/v}=-2u \space e^{2u/v}+2ve^{2u/v}$
