As with the previous problem we can convert the integral equation into a differential equation by taking the second derivative of the integral equation with respect to $x$. We find $y'' - \lambda y = 0.$ We immediately throw away the solution for $\lambda =0$ ($y = A x + B$) since it implies $y = 0$ in the integral equation. Thus, the solutions will be of the form $y = A \cosh\sqrt\lambda x + B \sinh\sqrt\lambda x.$ In fact, by examining the integral equation and its first derivative evaluated at $x=0$ and $x=1$ we can convince ourselves that the solutions must satisfy Robin boundary conditions $\begin{eqnarray*} y'(0) + y(0) &=& 0 \\ y'(1) - y(1) &=& 0. \end{eqnarray*}$ These boundary conditions make finding a closed form for the eigenvalues impossible. The solutions are peculiar. For $\lambda>0$ there is one eigenfunction. For \lambda<0 there is a tower of eigenfunctions. For large and negative $\lambda$ we will find approximate eigenvalues of the form $\lambda_n \approx -n^2\pi^2$.
The boundary conditions imply that $B = -A/\sqrt\lambda$ and that the eigenvalues satisfy the condition $\begin{equation} \tanh\sqrt\lambda = \frac{2\sqrt\lambda}{1+\lambda}. \tag{1} \end{equation}$
Case I: $\lambda > 0$
There is one solution to equation (1) for $\lambda>0$. It must be found numerically. It is $\lambda_0 \approx 2.38.$ The eigenfunction is $y_0 = A(\sqrt\lambda_0 \cosh \sqrt\lambda_0 x - \sinh\sqrt\lambda_0 x).$
Case II: \lambda < 0
Define $\mu = -\lambda$. The condition on the eigenvalues becomes $\begin{equation} \tan\sqrt\mu = \frac{2\sqrt\mu}{1-\mu}. \tag{2} \end{equation}$ There is an infinite tower of countable solutions to equation (2). We find, for example, $\mu_1 \approx 5.43 \approx \pi^2, \hspace{5ex} \mu_2 \approx 35.4 \approx (2\pi)^2, \hspace{5ex} \mu_3 \approx 84.8 \approx (3\pi)^2.$ In the limit of large $\mu$, the right-hand side of (2) vanishes. Thus, for large $\mu$, $\sqrt\mu = n \pi$ will be an approximate solution, where $n\in\mathbb{N}$. (These are the positive zeros of the tangent function.) That is, $\mu_n \approx n^2\pi^2$ for $n$ large. The eigenfunctions are
$y_n = A(\sqrt\mu_n \cos \sqrt\mu_n x - \sin\sqrt\mu_n x).$