The given differential equation is:

$$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$$

Dividing both sides by $x \log x$ (assuming $x>0$, $x \neq 1$), we obtain:

$$\frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^2}$$

This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \frac{1}{x \log x}$ and $Q(x) = \frac{2}{x^2}$.

The integrating factor (I.F.) is:

$$I.F. = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log x} \, dx}$$

Substituting $u = \log x$, $du = \frac{1}{x}dx$, we get:

$$\int \frac{1}{x \log x} dx = \int \frac{1}{u} du = \log |u| = \log |\log x|$$

Hence, $I.F. = e^{\log |\log x|} = |\log x|$. Taking the positive value, $I.F. = \log x$.

The solution of a linear differential equation is given by:

$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx + C$$

Substituting the values:

$$y \log x = \int \frac{2}{x^2} \cdot \log x \, dx + C$$

We evaluate $\int \frac{2 \log x}{x^2} dx$ using integration by parts.

Let $I = \int \frac{\log x}{x^2} dx$. Set $u = \log x$, $dv = x^{-2}dx$. Then $du = \frac{1}{x}dx$, $v = -x^{-1}$.

$$I = uv - \int v \, du = -\frac{\log x}{x} - \int \left(-\frac{1}{x}\right)\frac{1}{x}dx = -\frac{\log x}{x} + \int \frac{1}{x^2}dx = -\frac{\log x}{x} - \frac{1}{x}$$

Thus,

$$\int \frac{2 \log x}{x^2} dx = 2\left(-\frac{\log x}{x} - \frac{1}{x}\right) = -\frac{2}{x}(\log x + 1)$$

Therefore, the general solution is:

$$y \log x = -\frac{2}{x}(\log x + 1) + C$$

or equivalently,

$$y \log x + \frac{2}{x}(\log x + 1) = C$$