$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y(x)=k^2$$
$$\frac{dy}{dx}=\pm \sqrt{\frac{k^2}{y}-1}$$
ODE of the separable kind :
$$dx=\pm \sqrt{\frac{k^2-y^2}{y}}\:dy \quad\to\quad x=\pm\int \sqrt{\frac{k^2-y^2}{y}}\:dy $$
$$\pm x=k^2\tan^{-1}\left(\sqrt{\frac{y}{k^2-y}}\right)-\sqrt{(k^2-y)y}\:+c$$
The solution is on the form of $x$ as a function of $y$. There is no simple closed form for the inverse function $y(x)$.
The result can be presented on parametric form , with $\quad\tan(\theta)=\sqrt{\frac{y}{k^2-y}}$ :
$$\begin{cases}
x=k^2\left(\theta -\sin(\theta)\cos(\theta)\right)+c \\
y=k^2\sin^2(\theta)
\end{cases}$$
which is a parametric form of equation of cycloid.
$$ y''+ a[\sin(y)]+ b[\cos(y)] =f(x) \tag 1$$
$a[\sin(y)]+ b[\cos(y)]=\rho \sin(y+\phi)\quad\text{with}\quad \begin{cases}\rho=\sqrt{a^2+b^2}\\ \tan(\phi)=\frac{b}{a} \end{cases}$
Change of function :$\quad y(x)=u(x)-\phi\quad$ transform Eq.(1) into Eq.(2) :
$$u''+\rho\sin(u)=f(x) \tag 2$$
Change of variable :$\quad t=\sqrt{\rho}\:x\quad\to\quad \frac{d^2u}{dt^2}+\sin(u)=\frac{1}{\rho}f\left( \frac{t}{\sqrt{\rho}}\right)$
Let $\quad F(t)=\frac{1}{\rho}f\left( \frac{t}{\sqrt{\rho}}\right).\quad$ Since $\rho$ and $f$ are known, $F(t)$ is a known function.
$$\frac{d^2u}{dt^2}+\sin\left(u(t)\right)=F(t) \tag 3$$
Probably, no simpler form of equation (with no parameter inside) can be derived.
In the particular case $F(t)=0$ the solution can be expressed in terms of Jacobi elliptic function.
In the case $F(t)=C\neq 0$, Eq.(3) is an ODE of the autonomous kind. But the integration cannot be done in term of standard special functions.
A fortiori, in the general case $F(t)$ not constant, there is no closed form for the solution of the ODE.
This doesn't mean that closed form never exists in some particular cases. Just look backwards : Put a given function $y(x)$ into Eq.(1). This gives a particular function $f(x)$. For this function $f(x)$ , at least a closed form solution exist for Eq.(1) : the $y(x)$ a-priori chosen.
Thus, no definitive answer can be given to the question raised. This depends on the explicit definition of the function $f(x)$. But it is clear that analytical solving Eq.(1) is not possible in general.
Best Answer
You can try this $\displaystyle\frac{1}{xy}$.