Answer
$$\frac{dy}{dx} = \frac{y(y^2-3x^2)}{x(x^2-3y^2)}$$
Work Step by Step
Step-1: Differentiate the following equation with respect to $x$,
$$x^3y-xy^3=4$$, we get,
$$3x^2y+x^3\frac{dy}{dx}-y^3-3xy^2\frac{dy}{dx}=0$$
Step-2: Isolate $\frac{dy}{dx}$ terms together,
$$\frac{dy}{dx}(x^3-3xy^2)=y^3-3x^2y$$
$$\implies \frac{dy}{dx} = \frac{y(y^2-3x^2)}{x(x^2-3y^2)}$$
