We find
$$y(x) = \lambda \left[ (1-x)\int_0^x dt\, t y(t) + x\int_x^1 (1-t) y(t)\right].$$
This has a form similar to a Volterra equation of the first kind.
A standard technique is to take derivatives, thus transforming the integral equation into a differential equation.
Taking the second derivative of both sides with respect to $x$ we find
$$y'' = -\lambda y.$$
Thus, the solutions should be of the form
$$y = A \sin\sqrt{\lambda} x + B \cos\sqrt{\lambda} x.$$
Plugging this back into the original integral equation we find $B = 0$ and $\lambda = n^2 \pi^2$, where $n\in\mathbb{N}$.
Thus, the solutions are of the form
$$y = A \sin n \pi x$$
with eigenvalues
$$\lambda = n^2 \pi^2.$$
Addendum.
Another possible solution to the differential equation that we ignored above is $\lambda=0$ and $y = A+ B t$.
However, this is not compatible with the integral equation unless $A = B = 0$.
Notice also that hyperbolic solutions have been ruled out.
Initially we just assumed $\lambda$ was some complex number.
The integral equation then told us that $\lambda$ is real.