Example 5. $x\left(1+\left(y^{\prime}\right)^{2}\right)=1$.


Solution. In fact, this equation can be solved for the derivative $y^{\prime}: y^{\prime}= \pm \sqrt{\frac{1}{x}-1}$, and from this, we can obtain the general solution of the original equation. However, our goal is to illustrate the reasoning of point 6.3. Let $y^{\prime}=t, \quad \frac{d y}{d x}=t$, i.e., $d x=\frac{d y}{t}$. The original equation can be rewritten as $x=\frac{1}{1+t^{2}}$. Differentiating this equality with respect to $t: d x=-\frac{2 t}{\left(1+t^{2}\right)^{2}} d t$. Since $d x=\frac{d y}{t}$, we obtain the equation $d y=-\frac{2 t^{2}}{\left(1+t^{2}\right)^{2}} d t$ for the unknown function $y=y(x)=y(x(t))$. Integration gives: $y=-\operatorname{arctg} t+\frac{t}{t^{2}+1}+C$.

Combining this with the initial equality $x=\frac{1}{1+t^{2}}$, we thus obtain the general solution of the original equation in parametric form.

$$
\text { Answer. }\left\{\begin{array}{l}
x=\frac{1}{1+t^{2}} \\
y=-\operatorname{arctg} t+\frac{t}{t^{2}+1}+C
\end{array}\right.
$$