Answer
$$y = 6 \pi x-\frac{9 \pi^{2}}{2}$$
Work Step by Step
Given $$y=x^{2}(1-\sin x), \quad x=\frac{3 \pi}{2}$$
Since $y(3\pi/2)= \frac{9 \pi^{2}}{2}$ and
\begin{align*}
y'&=\frac{d}{dx}\left(x^2\right)\left(1-\sin \left(x\right)\right)+\frac{d}{dx}\left(1-\sin \left(x\right)\right)x^2 \\
&= 2x\left(1-\sin \left(x\right)\right)-x^2\cos \left(x\right)\\
y'\bigg|_{x=3\pi/2}&= 6\pi
\end{align*}
Then the tangent line equation is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-\frac{9 \pi^{2}}{2}}{x-3\pi/2} &= 6\pi \\
y- \frac{9 \pi^{2}}{2}&=6\pi [x-3\pi/2]\\
y&= 6 \pi x-\frac{9 \pi^{2}}{2}
\end{align*}
