I suppose that you meant $f:\mathbb{R}^2\rightarrow\mathbb{R}\ , f=f(x,y)$ (I am positive that you mean that since the question is if $f\in\mathcal{L}^1(\lambda_2)$, that is if is a (Lebesgue) integrable function in $\mathbb{R}^2$). We have that:
$\int_0^1f(x,y)\lambda(dx)=\int_0^yf(x,y)\lambda(dx)+\int_y^1f(x,y)\lambda(dx)=\int_0^yy^{-2}\lambda(dx)+\int_y^1-x^{-2}\lambda(dx)=y^{-2}\cdot\int_0^y\lambda(dx)-\int_y^1\left(x^{-1}\right)'\lambda(dx)=y^{-2}(y-0)-(1-\frac{1}{y})=\frac{2}{y}$
Then: $\int_0^1\frac{2}{y}\lambda(dy)=2\int_0^1y^{-1}\lambda(dy)=2\int_0^1(\log y)'\lambda(dy)=2(\log1-\lim_{t\rightarrow0^+}\log t)=+\infty$
In the same way:
$\int_0^1f(x,y)\lambda(dy)=\int_0^xf(x,y)\lambda(dy)+\int_x^1f(x,y)\lambda(dy)=\int_0^x-x^2\lambda(dy)+\int_x^1y^{-2}\lambda(dy)=-x^2\int_0^x\lambda(dy)+\int_x^1(-y^{-1})'\lambda(dy)=-x^2(x-0)+(-1-(-\frac{1}{x}))=-x^3+\frac{1}{x}-1$
and
$\int_0^1(-x^3+\frac{1}{x}-1)\lambda(dx)=-\int_0^1x^3\lambda(dx)+\int_0^1\frac{1}{x}\lambda(dx)-\int_0^1\lambda(dx)=-\int_0^1\left(\frac{x^4}{4}\right)'\lambda(dx)+\int_0^1(\log x)'\lambda(dx)-1=-(\frac{1}{4}-0)+(\log1-\lim_{t\rightarrow0^+}\log t)-1=-\frac{1}{4}-1+\infty=+\infty$
Since the integrals $\int_0^1\left(\int_0^1f(x,y)\lambda(dx)\right)\lambda(dy)=+\infty$ and $\int_0^1\left(\int_0^1f(x,y)\lambda(dy)\right)\lambda(dx)=+\infty$, we have that f is not integrable, that is $f\notin \mathcal L^1(\lambda_2)$