Answer
$F = (\frac{2\alpha}{x^3})~\hat{i}+(\frac{2\alpha}{y^3})~\hat{j}$
Work Step by Step
$U(x.y) = \alpha[(\frac{1}{x^2})+(\frac{1}{y^2})]$
$F = -((\frac{dU}{dx})~\hat{i}+(\frac{dU}{dy})~\hat{j})$
$F = -\alpha((\frac{-2}{x^3})~\hat{i}+(\frac{-2}{y^3})~\hat{j})$
$F = (\frac{2\alpha}{x^3})~\hat{i}+(\frac{2\alpha}{y^3})~\hat{j}$
