Answer
$$
\frac{y}{18}(x^{2}-64)+x^{\frac{2}{3}}y^{\frac{1}{3}}=12; \quad\quad x=8
$$
The equation of the tangent line at the given point $(8,27)$ is:
$$
y =\frac{-675}{4} x+1377
$$
Work Step by Step
$$
\frac{y}{18}(x^{2}-64)+x^{\frac{2}{3}}y^{\frac{1}{3}}=12; \quad\quad x=8
$$
find the$ y$-value :
$$
\begin{aligned}
\frac{y}{18}((8)^{2}-64)+(8)^{\frac{2}{3}}y^{\frac{1}{3}}&=12\\
\frac{y}{18}(64-64)+(2^{3})^{\frac{2}{3}}y^{\frac{1}{3}}&=12\\
0+4y^{\frac{1}{3}}&=12\\
y^{\frac{1}{3}}&=\frac{12}{4}=3\\
y &=3^{3}\\
\Rightarrow\quad\quad\quad\quad\quad\\
y &=27\\
\end{aligned}
$$
So, the given point is $(8,27)$ .
Now, we can calculate $\frac{dy}{dx}$ by implicit differentiation, use the product rule and the chain rule as follows:
$$
\begin{aligned}
\frac{y}{18}(2 x)+\frac{1}{18}\left(x^{2}-64\right) \frac{d y}{d x}+\frac{1}{3} x^{2 / 3} y^{-2 / 3} \frac{d y}{d x} & \\+\frac{2}{3} x^{-1 / 3} y^{1 / 3}=0 \\
\left(\frac{x^{2}-64}{18}+\frac{x^{2 / 3} y^{-2 / 3}}{3}\right) \frac{d y}{d x}=\frac{-2 x y}{18}-\frac{2 x^{-1 / 3} y^{1 / 3}}{3}& \\
\frac{d y}{d x}=\frac{\frac{-2 x y}{18}-\frac{2 x^{-1 / 3} y^{1 / 3}}{3}}{\frac{x^{2}-64}{18}+\frac{x^{2 / 3} y^{-2 / 3}}{3}} &\\
=\frac{-2 x y-12 x^{-1 / 3} y^{1 / 3}}{x^{2}-64+6 x^{2 / 3} y^{-2 / 3}}.
\end{aligned}
$$
To find the slope of the tangent line at the point $(8,27)$, let $x=8$ and $y=27.$ The slope is
$$
\begin{aligned}
\frac{-2(8)(27)-12(8)^{-1 / 3}(27)^{-1 / 3}}{64-64+6(8)^{2 / 3}(27)^{-2 / 3}} &=\frac{-432-18}{24} \\ &=(-450)\left(\frac{9}{24}\right) \\ &=\frac{-675}{4}.
\end{aligned}
$$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$$
\begin{aligned}
(y-y_{1}) &=m(x-x_{1})\\
(y-(27)) &=(\frac{-675}{4} )(x-(8))\\
y-27 &=\frac{-675}{4} x+1377 \\
y &=\frac{-675}{4} x+1377. \\
\end{aligned}
$$
Therefore, the equation of the tangent line at the given point $(8,27)$ is:
$$
y =\frac{-675}{4} x+1377.
$$
