# [Math] Equation of the tangent to $4x^2-3xy-y^2=25$ at $(2, -3)$

calculusimplicit-differentiationpolynomials

I have a function $f(x,y)=4x^2-3xy-y^2=25$ and I am trying to find the gradient at $(2,-3)$. First i took the derivative: $$\frac{d}{dx}(f'(x,y))=8x-\left(3y+3x\frac{dy}{dx}\right)-2y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{-8x+3y}{-3x-2y}=\frac{-16-9}{{-6+6}}=\frac{-25}{0}$$

What is going wrong here? After I get the gradient right, do I put it in $y-y_1=m(x-x_1)$, where $m$ is the gradient?

To see this more rigorously, you could consider $x$ as a function of $y$ and calculate $\frac{dx}{dy}$ at $(2,-3)$. You should get $0$. Then use this 'gradient' in the equation $m(y-y_1)=x-x_1$.