Dividing both sides by $x \log x$, we can rewrite the given equation as $\large\frac{dy}{dx}$$ + \large\frac{y}{\log x}$ $ = \large\frac{2}{x^2}$

The integrating factor is $= e^{\int \frac{dx}{x \log x}} $$= e^{\log (\log x)}$$ = \log x$

Multiplying the equation by $\ \log x$ we get: $\log x \large\frac{dy}{dx}$$ + \large\frac{y}{x}$$ = \large\frac{2}{x^2}$$ \log x$

Integrating, we get: $y \log x = \large \int \frac{2}{x^2} $$ \log x dx + c $

$\Rightarrow y \log x = 2 \large \int $$ \log x \times x^{-2} dx + c$

Using integration by parts and solving, we get: $y \log x = \large\frac{-2}{x}$$ (1+x) + c$