Find the equation of the tangent to the curve for the following equation at the point $(2, -3)$ $$4x^2-3xy-y^2=25$$
$$\therefore8x-3x\frac{dy}{dx}-3y-2y\frac{dy}{dx}=0 $$
$$\therefore\frac{dy}{dx} = \frac{8x-3y}{2y+3x}=\frac{16+9}{-6+6}=\frac{25}{0}$$
This is obviously undefined; however, the answer to this question is given to be $5x-2y=16$, suggesting that the gradient at the point is $2.5$. What is wrong with my working out?