Answer
$g_{u}=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$
and
$g_{v}=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$
Work Step by Step
Given: $g(u,v)=\frac{u+2v}{u^{2}+v^{2}}$
Need to find first partial derivatives $g_{u}$ and $g_{v}$.
Differentiate the function with respect to $u$ keeping $v$ constant.
$g_{u}=\frac{1(u^{2}+v^{2})-2u(u+2v)}{(u^{2}+v^{2})^{2}}$
$=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$
Differentiate the function with respect to $v$ keeping $u$ constant.
$g_{v}=\frac{2(u^{2}+v^{2})-2v(u+2v)}{(u^{2}+v^{2})^{2}}$
$=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$
Hence,
$g_{u}=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$
and
$g_{v}=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$
