Answer
$$ y = -2x+2$$
Work Step by Step
Given $$x^{2}+\sin y=x y^{2}+1, \ \ \ \ (1,0) $$
Differentiate both sides with respect to $x$
\begin{align*}
\frac{d}{dx}(x^{2}+\sin y)&=\frac{d}{dx}(x y^{2}+1)\\
2 x+\cos y y^{\prime}&=y^{2}+x(2 y) y^{\prime}\\
y^{\prime}(2 x y-\cos y)&=2 x-y^{2} \\
y'&= \frac{2 x-y^{2}}{2 x y-\cos y}
\end{align*}
Then at $(1,0)$
\begin{align*}
m&= \frac{2 }{-1}\\
&=- 2
\end{align*}
Then the tangent line is given by
\begin{align*}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-0}{x-1}&=-2\\
y &= -2x+2
\end{align*}
