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The given equation is $y\log y\frac{dx}{dy}+x=\frac{2}{y}$. Divide the entire equation by $y\log y$ to bring it into the standard form $\frac{dx}{dy} + P(y)x = Q(y)$:
$$\frac{dx}{dy} + \frac{1}{y\log y}x = \frac{2}{y^2\log y}$$Here, $P(y) = \frac{1}{y\log y}$. The integrating factor is given by $IF = e^{\int P(y) dy}$:
$$IF = e^{\int \frac{1}{y\log y} dy} = e^{\log(\log y)} = \log y$$The solution is $x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$:
$$x \log y = \int \frac{2}{y^2\log y} \cdot \log y \, dy = \int \frac{2}{y^2} dy$$Integrating $\frac{2}{y^2}$ gives $-2y^{-1} + C$:
$$x \log y = -\frac{2}{y} + C$$Final Answer: x \log y = -\frac{2}{y} + C
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