Example 5. Solve the equation $\frac{\left(y^{\prime \prime}\right)^{2}-y^{\prime} \cdot y^{\prime \prime \prime}}{\left(y^{\prime}\right)^{2}}=\frac{1}{x^{2}}$.
Solution. Before solving a given equation, it makes sense to analyze its structure to identify integrable expressions or combinations.

In this case, we notice the equalities (derivatives are taken with respect to $x$):

$$
\frac{\left(y^{\prime \prime}\right)^{2}-y^{\prime} \cdot y^{\prime \prime \prime}}{\left(y^{\prime}\right)^{2}}=-\left(\frac{y^{\prime \prime}}{y^{\prime}}\right)^{\prime}, \quad \frac{1}{x^{2}}=-\left(\frac{1}{x}\right)^{\prime}
$$

This means the original equation can be rewritten as a separable equation (with entire expressions instead of variables):

$$
\left(\frac{y^{\prime \prime}}{y^{\prime}}\right)^{\prime}=\left(\frac{1}{x}\right)^{\prime}
$$

From this, we get the equality

$$
\frac{y^{\prime \prime}}{y^{\prime}}=\frac{1}{x}+C_{1}
$$

In the left part, we have $\left(\ln y^{\prime}\right)^{\prime}$, so after integration, we get $\ln \left|y^{\prime}\right|=\ln |x|+C_{1} x+\ln C_{2}$, i.e., $y^{\prime}=C_{2} x \cdot e^{C_{1} x}$. Integrating by parts, we have:

$$
y=C_{2} \int x e^{C_{1} x} d x=C_{2}\left(\frac{x}{C_{1}} e^{C_{1} x}-\frac{e^{C_{1} x}}{C_{1}^{2}}\right)+C_{3}
$$

Thus, the general solution is obtained.

Answer. $y=\frac{C_{2}}{C_{1}} e^{C_{1} x}\left(x-\frac{1}{C_{1}}\right)+C_{3}$.