Answer
$$
2y^{3}(x-3)+x\sqrt {y}=3; \quad\quad x=3
$$
The equation of the tangent line at the given point is:
$$
y =-2x+7
$$
Work Step by Step
$$
2y^{3}(x-3)+x\sqrt {y}=3; \quad\quad x=3
$$
find the$ y$-value :
$$
\begin{aligned}
2y^{3}((3)-3)+(3)\sqrt {y}&=3\\
2y^{3}(0)+(3)\sqrt {y}&=3\\
(3)\sqrt {y}&=3\\
\sqrt {y}&=1\\
\Rightarrow\quad\quad\quad\quad\quad\\
y &=1.\\
\end{aligned}
$$
So, the given point is $(3,1)$ .
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation, use the product rule and the chain rule as follows:
$$
\begin{aligned} 2 y^{3}(1)+6 y^{2}(x-3) \frac{d y}{d x} &
+x\left(\frac{1}{2}\right) y^{-1 / 2} \frac{d y}{d x}+\sqrt{y} &=0 \\
6 y^{2}(x-3) \frac{d y}{d x}+\frac{x}{2 \sqrt{y}} \frac{d y}{d x} &=-2 y^{3}-\sqrt{y} \\
\left(6 y^{2}(x-3)+\frac{x}{2 \sqrt{y}}\right) \frac{d y}{d x} &=-2 y^{3}-\sqrt{y} \\
\frac{d y}{d x} &=\frac{-2 y^{3}-\sqrt{y}}{6 y^{2}(x-3)+\frac{x}{2 \sqrt{y}}} \\
\frac{d y}{d x} &=\frac{-4 y^{7 / 2}-2 y}{12 y^{5 / 2}(x-3)+x} \end{aligned}
$$
To find the slope of the tangent line at the point $(3,1)$, let $x=3$ and $y=1.$ The slope is
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{-4 y^{7 / 2}-2 y}{12 y^{5 / 2}(x-3)+x} \\
&=\frac{-4 (1)^{7 / 2}-2 (1)}{12 (1)^{5 / 2}((3)-3)+(3)}\\
&=\frac{-6}{3}\\
&=-2.
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation
of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(1)) &=(-2)(x-(3))\\
y-1 &=-2x+6\\
y &=-2x+7\\
\end{aligned}
$$
Therefore, the equation of the tangent line at the given point is:
$$
y =-2x+7
$$
