A curve has equation $x^2 - 4xy + 2y^2 = 1$.
a) Find ad simplify an expression for $\frac{dy}{dx}$.
b) Show that the tangent to the curve at the point $P=(1,2)$ has the equation
$$3x - 2y +1 = 0.$$
The tangent to the curve at the point $Q$ is parallel to the tangent at $P$.
c) Find the coordinates of $Q$.
This is going by @JamesS.Cook's remarks above; If you happen to be familiar with the implicit function theorem, we could be a little more exact.
Hint: for simplicity's sake, suppose we know that we can represent $y=y(x)$ such that $(x,y(x))$ is always a (unique, in the appropriate sense) point on that curve (in fact, such a representation is only possible in a small neighborhood of each point on that curve, but that needn't hinder us here). If we then define $$z(x) = x^2 -4xy(x) + 2(y(x))^2,$$ what does that say about $\frac{dz}{dx}$, and what would the chain rule then imply regarding $\frac{dy}{dx}$?