Answer
$f_x=2/3(x^3+y/2)^{-1/3}\times 3x^2$
$f_y=2/3(x^3+y/2)^{-1/3}\times 1/2$
Work Step by Step
Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y as a constant, and vice versa:
$f_x=2/3(x^3+y/2)^{-1/3}\times 3x^2$
$f_y=2/3(x^3+y/2)^{-1/3}\times 1/2$