Answer
$f_x=2x^2 (x^3+\dfrac{y}{2})^{-1/3}$ and
$f_y=\dfrac{1}{3} (x^3+\dfrac{y}{2})^{-1/3}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ as a constant, and vice versa:
$f_x=\dfrac{2}{3} \times (x^3+y/2)^{-1/3}\times 3x^2$ and
$f_y=\dfrac{2}{3} \times (x^3+y/2)^{-1/3}\times 1/2$
Our results are: $f_x=2x^2 (x^3+\dfrac{y}{2})^{-1/3}$ and
$f_y=\dfrac{1}{3} (x^3+\dfrac{y}{2})^{-1/3}$
