Answer
$$y'=\frac{2-3x^2-4y}{4x-4y^{1/3}}$$
Work Step by Step
$$x^3+4xy-3y^{4/3}=2x$$
Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$3x^2+4(y+xy')-3\times\frac{4}{3}y^{1/3}\times y'=2$$
$$3x^2+4y+4xy'-4y^{1/3}y'=2$$
Now we set all elements with $y'$ to one side and the rest to the other side: $$4xy'-4y^{1/3}y'=2-3x^2-4y$$ $$y'(4x-4y^{1/3})=2-3x^2-4y$$
Finally, we calculate $y'$: $$y'=\frac{2-3x^2-4y}{4x-4y^{1/3}}$$
