Level curves and intersections Find an equation of the curve in the xy plane which passes through the point (1,1) and intersects all level curves of the function $f(x,y) = x^{4} + y^{2}$ at right angles.

I don't know where to start. I have realized that the level curves are when the value of $f(x,y)$ remains fixed, so we would have $k=x^{4}+y^{2}$. How can i continue from here? I tried to actually set up a differential equation (by finding the orthogonal families of curves to these level curves), but i'm not sure if this is the right way to do it.
Is there a general method (involving multivariable calculus) to prove that two curves or surfaces intersect at right angles?
 A: There are several ways to go about it.  Here is a multivariable calculus approach.
Level curves of $f(x,y) = x^4 + y^2$ will be perpendicular to the direction of the gradient of $f$, $$\nabla f = 4x^3\hat{\mathbf{x}} + 2y\hat{\mathbf{y}}.$$  Furthermore, level curves of $g = g(x,y)$ will be perpendicular to level curves of $f$ if $\nabla f$ and $\nabla g$ are everywhere orthogonal.  $\nabla f$ and $\nabla g$ are orthogonal if their dot product is zero:
\begin{align}
0 &= \nabla f\cdot\nabla g \\
&= 4x^3\frac{\partial g}{\partial x} + 2y\frac{\partial g}{\partial y}
\end{align}
This is a partial differential equation for the unknown function $g$.  This PDE is satisfied when
\begin{align}
\frac{\partial g}{\partial x} &= \frac{1}{4x^3} \\
\frac{\partial g}{\partial y} &= -\frac{1}{2y}
\end{align}
After integrating each, it is clear that, up to an arbitrary constant $g$ must be
\begin{align}
g(x,y) = -\frac{1}{8x^2} - \frac{1}{2}\ln(y)
\end{align}
All level curves of $g$ will be orthogonal to level curves of $f$; now simply choose the one where $g(1,1) = g(x,y)$.
