In a double integral, the outer limits must be constant, but the inner limits can depend on the outer variable. This means, we must put y as the inner integration variables.
$\int_{0}^{1}\int_{0}^{1/x}x/1+y^{2}dxdy$
$\int_{0}^{1}x\tan^{-1}(y)]_{0}^{1/x}dx$
$\int_{0}^{1}x(\tan^{-1}(1/x)-\tan^{-1}(0))dx$
$\int_{0}^{1}x\tan^{-1}(1/x)dx$
$[\tan^{-1}(1/x)\frac{x^{2}}{2}-1/2[ \int \frac{x^{2}}{x^{2}+1} x^{2} dx ]]_{0}^{1}$
$[\tan^{-1}(1/x)\frac{x^{2}}{2}-1/2[\frac{x^{3}}{3}-x+\tan^{-1}(x)]]_{0}^{1}$
$(\frac{1}{2}*\frac{\Pi }{4}-1/2[1/3-1+\frac{\Pi }{4}])$
$\frac{\Pi }{8}-1/2[-2/3+\frac{\Pi }{4}$
$\frac{\Pi }{8}+\frac{1}{3}-\frac{\Pi }{8}$
=1/3
Hence answer should be 1/3.