Answer
$v_x= \dfrac{\ln v}{\ln u \ln v -1}$
Work Step by Step
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$.
$1=v_x \ln u +(v) (u^{-1} ) u_x$ and $0=u_x \ln v +(u) (v^{-1} ) v_x$
So, $u_x= \dfrac{-(u/v) v_x}{\ln v}$
Now, $1=v_x \ln u +(v) (u^{-1} ) (u_x= \dfrac{-(u/v) v_x}{\ln v})$
$\implies v_x= \dfrac{\ln v}{\ln u \ln v -1}$
