Divide through by $x{\cos }^{2}y$:

${\sec }^{2}y\frac{dy}{dx}-2\tan y\cdot \frac{1}{x}=x^2(2-x^3)$ (using $\sin 2y=2\sin y\cos y$, so $\sin 2y/{\cos }^{2}y=2\tan y$).

Let $t=\tan y$, so $\frac{dt}{dx}={\sec }^{2}y\frac{dy}{dx}$. The equation becomes linear in $t$:

$\frac{dt}{dx}-\frac{2t}{x}=x^2(2-x^3)$.

Integrating factor: $\mathrm{IF}=e^{\int -2/xdx}=e^{-2\ln x}=\frac{1}{x^2}$.

$\frac{t}{x^2}=\int \frac{1}{x^2}\cdot x^2(2-x^3)dx=\int (2-x^3)dx=2x-\frac{x^4}{4}+C$.

Apply $y(2)=0\Rightarrow t=0$ at $x=2$: $0=4-4+C\Rightarrow C=0$.

$\therefore \tan y=2x^3-\frac{x^6}{4}$.

At $x=1$: $\tan y(1)=2-\frac{1}{4}=\frac{7}{4}$.

Option (2).