The gradient of $f(x,y) = x^2 + y^4$ is tangent to the curve $\gamma(t)=(t^2,t)$, at a point $P = \gamma(t_0)$, with $t_0 > 0$. Consider the level curve of $f$ that contains $P$. Find the equation of the tangent line to this curve at point $P$.

What I got from the question is:

1) The gradient vector is perpedincular to the direction vector of the line I'm trying to find;

2) The gradient vector is parallel to the derivative of $\gamma(t_0)$

But I think I'm missing something to continue.

First lets calculate a few things: $$\nabla f(x,y)=(2x,4y^3)$$ $$\gamma'(t)=(2t,1)$$ $$f(\gamma(t_0))=(t_0^2)^2+t_0^4=2t_0^4$$

Hence, the level curve of $f$ that contains $P$ is given by: $$\left \{(x,y)\in \mathbb{R}^2 : f(x,y)=2t_0^4 \right \}$$

The fact that $\nabla f(\gamma(t_0))$ is tangent to the curve $\gamma$ at $P=\gamma(t_0)$ means that the vectors $\nabla f(\gamma(t_0))$ and $\gamma'(t_0)$ are colinear, this means that for some constant $\lambda$:

$$\nabla f(\gamma(t_0))=\lambda \gamma'(t_0)$$

From the equations above this translates to:

\begin{cases} 2t_0^2=\lambda 2t_0 \\ 4t_0^3=\lambda \end{cases}

Substituting the second equation into the first we get:

$$2t_0^2=(4t_0^3)2t_0 \implies2t_0^2(4t_0^2-1)=0 \implies t_0=\frac{1}{2}$$

since $t_0>0$.

Can you take it from here?