Answer
$$-\frac{34}{65}$$
Work Step by Step
Given $$y^{2} x^{3}+y^{3} x^{4}-10 x+y=5, \ \ \ p( 2,1)$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{d x}\left(y^{2} x^{3}+y^{3} x^{4}-10 x+y\right)&=\frac{d}{d x}(5)\\
2 y x^{3} y^{\prime}+3 y^{2} x^{4} y^{\prime}+y^{\prime}-10+3 x^{2} y^{2}+4 x^{3} y^{3}&=0\\
\left(2 y x^{3}+3 y^{2} x^{4}+1\right) y^{\prime}&=10-3 x^{2} y^{2}-4 x^{3} y^{3}
\end{align*}
Then
\begin{aligned}
y^{\prime} &=\frac{\left(10-3 x^{2} y^{2}-4 x^{3} y^{3}\right)}{\left(2 y x^{3}+3 y^{2} x^{4}+1\right)} \\
\left.y^{\prime}\right|_{(2,1)} &=\frac{(10-12-32)}{(16+48+1)} \\
&=-\frac{34}{65}
\end{aligned}
