Answer
$$y = 2x+\pi$$
Work Step by Step
Given $$\sin (2 x-y)=\frac{x^{2}}{y}, \quad(0, \pi)$$
Differentiate both sides
\begin{align*}
\frac{d}{dx}( \sin (2 x-y))&=\frac{d}{dx}\left(\frac{x^{2}}{y}\right)\\
2 \cos (2 x-y)-y^{\prime} \cos (2 x-y)&=\frac{2 x}{y}-\frac{x^{2} y^{\prime}}{y^{2}}\\
\frac{x^{2}-y^{2} \cos (2 x-y)}{y^{2}} y^{\prime}&=\frac{2 x-2 y \cos (2 x-y)}{y}\\
y'&=\frac{2 y(x-y \cos (2 x-y))}{x^{2}-y^{2} \cos (2 x-y)}
\end{align*}
Then at $ (0,\pi)$
\begin{align*}
m&= \frac{2 \pi(0-\pi \cos (-\pi))}{0-\pi^{2} \cos (-\pi)}\\
&= 2
\end{align*}
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-\pi}{x-0}&=2\\
y-\pi&= 2x\\
y&= 2x+\pi
\end{align*}
