Spectrum of integral operator 
Given $g\in C^1([0,1]\times[0,1])$, consider the operator  $$Tu(x) =
 \int_0^1 g(x,t) u(t) dt$$ defined on $u\in C([0,1])$. Discuss the
  spectrum of T.

My attempt:
First I can show that $T$ is a compact operator. Given a sequence $u_n$ bounded in $C([0,1])$$\|u_n \|_\infty \leq M$, and let $\|g\|_\infty = N$, we have 
$$|(Tu_n)(x)| \leq M\cdot N$$
And given $x_0$ we have
$$|(Tu_n)(x_0) - (Tu_n)(x)| \leq \int_0^1 |g(x_0,t) - g(x,t)| M dt$$
by continuity of $g$, we see that $(Tu_n)(x)$ is equicontinuous. 
Now we know that $0$ is in the spectrum of $T$. To look for other eigenvalues, let $\lambda \neq 0$ and $f\neq 0$.
$$\lambda f(x) = \int_0^1 g(x,t)f(t) dt$$
using the Lebesgue differential theorem, I get that 
$$\lambda f'(x) = \frac{d}{dx} \int_0^1 g(x,t)f(t) dt = \int_0^1 \frac{\partial}{\partial x} g(x,t)f(t) dt.$$
How would I continue from here? 
Thank you very much!

In order to show that your operator is compact, show that it maps a bounded sequence $\{ f_{n} \}_{n=1}^{\infty}\subset C[0,1]$ to an equicontinuous sequence of functions. So, let $\{ f_{n} \}_{n=1}^{\infty}$ satisfy $\|f_{n}\|_{C[0,1]}\le M$ for all $n$ and some fixed $M$; then, for every $\epsilon > 0$, show that there is a $\delta > 0$ such that
$$
                        |Tf_{n}(x)-Tf_{n}(y)| < \epsilon
$$
holds for all $n$ whenever $|x-y| < \delta$. Differentiability is a stronger condition than what you need, but it makes the proof a little easier:
$$
\begin{align}
         |Tf_{n}(x)-Tf_{n}(y)| & \le \int_{0}^{1}|g(x,t)-g(y,t)||f_{n}(t)|\,dt \\
                     & \le M \int_{0}^{1}|g(x,t)-g(y,t)|\,dt \\
                     & \le M \int_{0}^{1}\left|\int_{x}^{y}\frac{\partial g}{\partial u}g(u,t)\,du\right|\,dt.
\end{align}
$$
Let $L$ be a bound for $\frac{\partial g}{\partial u}$ on $[0,1]\times[0,1]$, and you get
$$
                          |Tf_{n}(x)-Tf_{n}(y)| \le LM|x-y|.
$$
For the $\epsilon > 0$ which was given, choose $\delta = \frac{\epsilon}{2LM}$ and you get the desired equicontinuity of $\{ f_{n} \}_{n=1}^{\infty}$, i.e., that $|Tf_{n}(x)-Tf_{n}(y)| < \epsilon$ for all $n$ whenever $|x-y| < \delta$.
As you suggest, once you know $T$ is compact, you should be able to say more about the spectrum.
