Questions: If -x^2 + y^3 = -xy - 1 then find dy/dx in terms of x and y.

Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) for the equation \(-x^{2} + y^{3} = -xy - 1\).

Differentiate both sides of the equation.

Differentiating the left side gives \(-2x + 3y^{2} \frac{dy}{dx}\), and the right side gives \(-y - x \frac{dy}{dx}\).

Set the derivatives equal to each other.

We have \(-2x + 3y^{2} \frac{dy}{dx} = -y - x \frac{dy}{dx}\).

Rearrange to solve for \(\frac{dy}{dx}\).

This leads to \((3y^{2} + x) \frac{dy}{dx} = -y + 2x\), thus \(\frac{dy}{dx} = \frac{-y + 2x}{3y^{2} + x}\).

The final answer is \(\boxed{\frac{-y + 2x}{3y^{2} + x}}\).

The answer is \(\boxed{\frac{-y + 2x}{3y^{2} + x}}\).