Answer
$$\frac{dy}{dx}=-\frac{2(x+2y)}{4x-3y^2}$$
Work Step by Step
Step-1: Differentiate the following equation:
$$x^2 + 4xy - y^3=6$$, we get,
$$2x + 4y + 4x\frac{dy}{dx} - 3y^2\frac{dy}{dx}=0$$
Step-2: Isolate $\frac{dy}{dx}$ terms together,
$$\frac{dy}{dx}(4x-3y^2) = -2(x+2y)$$
$$\implies \frac{dy}{dx} = -\frac{2(x+2y)}{4x-3y^2}$$
