Implicit differentiation to find dy/dx

Differentiate both sides of $x^2 - xy + y^2 = 4$ with respect to $x$. For the term $x^2$ we get $2x$. For the term $xy$, apply the product rule: $\frac{d}{dx}[xy] = (1)y + x\frac{dy}{dx} = y + x\frac{dy}{dx}$. For $y^2$, apply the chain rule: $\frac{d}{dx}[y^2] = 2y\frac{dy}{dx}$. The right side, $4$, differentiates to $0$. This gives $2x - \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0.$

Expand to get $2x - y - x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$. Group the $\frac{dy}{dx}$ terms: $(2y - x)\frac{dy}{dx} = y - 2x$. Dividing both sides by $(2y-x)$ gives $\frac{dy}{dx} = \frac{y - 2x}{2y - x}.$