Problem

A curve has equation

$$y^2=ye^{-2x}-3x$$

(a) Show that

$$\frac{dy}{dx}=\frac{2ye^{-2x}+3}{e^{-2x}-2y}$$

The curve crosses the $y$-axis at the origin and at the point $P$.

The tangent to the curve at the origin and the tangent to the curve at $P$ meet at the point $R$.

(b) Find the coordinates of $R$.