Answer
The equation of the tangent line is $y=\dfrac{4x}{5}+\dfrac{12}{15}$.
Work Step by Step
Firstly differentiate $x^3 + y^3 = 3xy$ with respect to $x$ using product rule and chain rule.
We get, $3x^2+3y^2\dfrac{dy}{dx}=3x\dfrac{dy}{dx}+3y\dfrac{dx}{dx}$
$\implies 3x^2+3y^2\dfrac{dy}{dx}=3x\dfrac{dy}{dx}+3y$
Now solve for $\dfrac{dy}{dx}$ as follows.
$3x^2+3y^2\dfrac{dy}{dx}=3x\dfrac{dy}{dx}+3y$
$\implies 3y^2\dfrac{dy}{dx}-3x\dfrac{dy}{dx}=3y-3x^2$
$\implies (3y^2-3x)\dfrac{dy}{dx}=3y-3x^2$
$\implies \dfrac{dy}{dx}=\dfrac{3y-3x^2}{3y^2-3x}$
Now substitute $x=\dfrac{2}{3}$ and $y=\dfrac{4}{3}$ in the above equation to find the slope of the tangent line at $\left(\dfrac{2}{3},\dfrac{4}{3}\right)$.
$\dfrac{dy}{dx}=\dfrac{3\times\left(\dfrac{4}{3}\right)-3\times\left(\dfrac{2}{3}\right)^2}{3\times\left(\dfrac{4}{3}\right)^2-3\times\left(\dfrac{2}{3}\right)}=\dfrac{3\times\left(\dfrac{4}{3}\right)-\left(\dfrac{4}{3}\right)}{\left(\dfrac{16}{3}\right)-\left(\dfrac{6}{3}\right)}=\dfrac{2\times\left(\dfrac{4}{3}\right)}{\left(\dfrac{10}{3}\right)}=\dfrac{\left(\dfrac{8}{3}\right)}{\left(\dfrac{10}{3}\right)}\hspace{24px}=\dfrac{8}{10}=\dfrac{4}{5}$
Now use the slope-intercept form of the tangent line.
That is, $y-\dfrac{4}{3}=\dfrac{4}{5}\left(x-\dfrac{2}{3}\right)$
Now simplify to get the equation of the tangent line at $\left(\dfrac{2}{3},\dfrac{4}{3}\right)$.
We get, $y-\dfrac{4}{3}=\dfrac{4x}{5}-\dfrac{8}{15}$
$\implies y=\dfrac{4x}{5}-\dfrac{8}{15}+\dfrac{4}{3}$
$\implies y=\dfrac{4x}{5}+\dfrac{12}{15}$
Hence, the equation of the tangent line is $y=\dfrac{4x}{5}+\dfrac{12}{15}$.
