Answer
$$y'=\frac{-1}{4}$$
Work Step by Step
Given $$\sin ^{2}(3 y)=x+y, \quad\left(\frac{2-\pi}{4}, \frac{\pi}{4}\right)$$
Differentiate with respect to $x$
\begin{align*}
2\sin(3 y)\cos(3y)y'&=1+y'\\
( \sin(6y)-1)y'&= 1
\end{align*}
Then
\begin{aligned}
y'&= \frac{1}{\sin(6y)-1}\\
\left.y^{\prime}\right|_{\left(\dfrac{2-\pi}{4}, \dfrac{\pi}{4}\right)} &=\frac{1}{\sin(3\pi/2)-1} \\
&= \frac{-1}{4}
\end{aligned}
