## Calculus (3rd Edition)

$f_{xv}=0$
We first differentiate with respect to $x$; then we have $$f_x=\frac{2x}{3y^2+\ln(2+u^2)}.$$ Then, we compute $f_{xv}$, which gives directly $$f_{xv}=0.$$ Now, we have $$f_{uvxyvu}=0.$$