Answer
$y=\frac{7\sqrt{2}}{8}x-\frac{3\sqrt{2}}{4}$
Work Step by Step
$y^2(6-x)=x^3$ (Take the implicit differentiation)
$\frac{d}{dx}(y^2(6-x))=\frac{d}{dx}(x^3)$ (Use the Product Rule)
$\frac{d}{dx}(y^2)\cdot (6-x)+y^2\cdot \frac{d}{dx}(6-x)=3x^2$
$2y\cdot \frac{dy}{dx}\cdot (6-x)+y^2(-1)=3x^2$ (Find $\frac{dy}{dx}$ at $(x,y)=(2,\sqrt{2})$
$2\sqrt{2}\cdot \frac{dy}{dx}\cdot (6-2)+\sqrt{2}^2(-1)=3\cdot 2^2$
$8\sqrt{2}\frac{dy}{dx}-2=12$
$8\sqrt{2}\frac{dy}{dx}=14$
$\frac{dy}{dx}=\frac{7}{4\sqrt{2}}$
So, the slope of the tangent line to the curve at $(2,\sqrt{2})$ is $m=\frac{7}{4\sqrt{2}}$.
Now, find the equation of the tangent line:
$y-\sqrt{2}=\frac{7}{4\sqrt{2}}(x-2)$
$y-\sqrt{2}=\frac{7}{4\sqrt{2}}x-\frac{7\sqrt{2}}{4}$
$y=\frac{7}{4\sqrt{2}}x-\frac{3\sqrt{2}}{4}$
$y=\frac{7\sqrt{2}}{8}x-\frac{3\sqrt{2}}{4}$
