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.