$2xp=2\tan y+p^{3}\cos^{2}y$ ($p=\frac{dy}{dx}$)
Multiply both sides by $\cos y$, $2xp\cos y=2\sin y+p^{3}\cos ^{3}y$
Let $t=p\cos y$,the equation becomes$$ \sin y=xt-\frac{1}{2}t^{3} $$ Take the derivative of both sides with respect to $x$, $$ t=t+x \frac{dt}{dx}-\frac{3}{2}t^{2} \frac{dt}{dx} $$ $$ \left( x-\frac{3}{2}t^{2} \right) \frac{dt}{dx}=0 $$ I don't know what to do next,$x- \frac{3}{2}t^{2}=0$ and $\frac{dt}{dx}=0$ still equation in terms of $x,y$ and $\frac{dy}{dx}(p)$.
Your transformations are very suggestive leading to consider $z=\sin y$. Then the equation becomes, with $z'=\cos(y)y'=t$, $$ z=xz'-\frac12(z')^3=xt-\frac12t^3. $$ This now has the form of a Clairaut DE, with its immediate solution by lines and the envelope.
Let's first solve $\frac{dt}{dx}=0$: $$ \frac{dt}{dx}=0 \implies t=C_1 \implies \frac{dy}{dx}\cos y= C_1 \implies \sin y=C_1x+C_2. \tag{1} $$ Since the original ODE is of first order, we expect that its general solution has a single constant of integration, whereas $(1)$ has two. To determine the relation beween $C_1$ and $C_2$, we substitute $(1)$ in the original ODE: \begin{align} 2x\frac{C_1}{\cos y}=2\tan y+\frac{C_1^3}{\cos^3 y}\cos^2y \implies 2C_1x=2\sin y+C_1^3 \\ \implies 2C_1x=2(C_1x+C_2)+C_1^3 \implies C_2=-\frac{C_1^3}{2}. \tag{2} \end{align} Therefore, the general solution to the original ODE is $$ \sin y=C_1x-\frac{C_1^3}{2}. \tag{3} $$
Next, instead of solving $x-\frac{3}{2}t^2=0$, we notice that the envelope of $(3)$ is also a solution to the original ODE. To find it, eliminate $C_1$ between $(3)$ and $$ \frac{\partial}{\partial C_1}\left\{\sin y-C_1x+\frac{C_1^3}{2}\right\}=0 \implies x-\frac{3C_1^2}{2}=0. \tag{4} $$ Solving $(4)$ for $C_1$ and plugging the result into $(3)$, we obtain $$ \sin y=\pm\left(\frac{2x}{3}\right)^{3/2}\qquad\left(0\leq x\leq \frac{3}{2}\right). \tag{5} $$ It's straightforward to show that $(5)$ is a solution to $x-\frac{3}{2}t^2=0$.
