Consider the curve defined by x^2 + 4y^2 = 7 + 3xy
Show that there is a point P with x-coordinate at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P
dy/dx = (3y - 2x)/ (8y-3x)
I think you forgot to tell us what the x-coordinate is. For now I'll just call it $\displaystyle x_0$.
So the derivative is $\displaystyle \frac{dy}{dx}=\frac{3y-2x}{8y-3x}$. The horizontal tangent line occurs when the derivative is zero, namely, when $\displaystyle 3y-2x=0$. Since we are given the x-coordinate (which I'm calling $\displaystyle x_0$), we have that $\displaystyle 3y-2x_0=0$. Solving for $\displaystyle y$ gives us $\displaystyle y=\frac{2x_0}{3}$ so the point is $\displaystyle P=\left(x_0,\frac{2x_0}{3}\right)$.
So whatever the x-coordinate actually is, just plug it in for $\displaystyle x_0$ and that will give you the answer.