Let $\lambda$ be a non-zero eigenvalue. By definition, there exists a non-zero function $f \in L^2[0,1]$ such that
$$T(f)=\lambda f$$
i.e. such that for every $x \in [0,1]$,
$$\int_0^1(2xy-x-y+1)f(y)dy = \lambda f(x) $$
i.e. $$x\int_0^1(2y-1) f(y) dy + \int_0^1 (1-y)f(y)dy = \lambda f(x) \quad \quad (*)$$
so there exists $a,b \in \mathbb{R}$ such that $f(x)=ax+b$. Replacing in $(*)$, one gets $$x\int_0^1(2y-1)(ay+b) dy + \int_0^1 (1-y)(ay+b)dy = \lambda (ax+b)$$
i.e. $$x \times \dfrac{a}{6} + \dfrac{a+3b}{6} = \lambda ax+\lambda b$$
so identifying the coefficients, you get $\dfrac{a}{6}=\lambda a$ and $\dfrac{a+3b}{6}=\lambda b$, i.e. (since $a$ and $b$ cannot be zero at the same time), $\lambda = \dfrac{1}{6}$ or $\lambda=\dfrac{1}{2}$.
So the only possible eigenvalues of $T$ are $0$, $\dfrac{1}{2}$ and $\dfrac{1}{6}$. It remains to see why they are really eigenvalues.
For $0$, let $f(x)=6x^2-6x+1$ : by direct calculation, one has $Tf=0$, so $f$ is an eigenvector associated to the eigenvalue $0$.
For $1/2$, let $f(x)=1$ : by direct calculation, one has $Tf(x)=\dfrac{1}{2}=\dfrac{1}{2}f(x)$, so $f$ is an eigenvector associated to the eigenvalue $1/2$.
For $1/6$, let $f(x)=2x-1$ : by direct calculation, one has $Tf(x)=\dfrac{x}{3}-\dfrac{1}{6} = \dfrac{2x-1}{6} = \dfrac{1}{6}f(x)$, so $f$ is an eigenvector associated to the eigenvalue $1/6$.
Finally, the eigenvalues of $T$ are exactly $0$, $\dfrac{1}{2}$ and $\dfrac{1}{6}$.
