Solution
= Solution
For $f(u,v)=\ln u+\ln v$, the needed <partial derivatives> are
$$
f_v=\frac1v,\qquad f_{uu}=-\frac1{u^2},\qquad f_{uv}=0,\qquad f_{vv}=-\frac1{v^2}.
$$
Substituting these into part (ii) gives
$$
f_{xy}=-\frac{4xy}{u^2}-\frac{4x^3y^3}{v^2}+\frac{4xy}{v}.
$$
Since $v-x^2y^2=1$, the last two terms combine. Thus
$$
\boxed{f_{xy}=4xy\left[\frac1{(1+x^2y^2)^2}-\frac1{(1+x^2+y^2)^2}\right].}
$$