Answer
$y(x)=\arcsin(C_1x+C_2)$
Work Step by Step
We are given:
$$y''=(y')^2\tan y$$
Suppose that $a= \dfrac{dy}{dx}$ and $ \dfrac{da}{dx}= \dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}\dfrac{dx}{dy}=a\dfrac{da}{dy}$
Now, $a\dfrac{da}{dy}=a^2\tan y$
$\dfrac{da}{dy}=a\tan y$
$\frac{da}{a}=\tan y dy$
Integrate to obtain: $\ln a=-\ln (\cos x)+C$
$ \implies a=C\sec(y)$
$\implies \frac{dy}{dx}=C\sec(y)$
$\implies \cos (y)dy=Cdx$
Continue to integrate:
$\sin y=C_1x+C_2$
Therefore, the general solution is: $y(x)=\arcsin(C_1x+C_2)$
