use implicit differentiation to find an equation of the tangent line to the curve at the given point…

Explanation:

Step1: Differentiate the first equation implicitly

Differentiate $x^{\frac{2}{3}}+y^{\frac{2}{3}} = 4$ with respect to $x$. Using the power - rule and chain - rule, we have $\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}y'=0$.

Step2: Solve for $y'$

First, factor out $\frac{2}{3}$: $\frac{2}{3}(x^{-\frac{1}{3}}+y^{-\frac{1}{3}}y') = 0$. Then, $x^{-\frac{1}{3}}+y^{-\frac{1}{3}}y'=0$. Rearrange to get $y'=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$.

Step3: Evaluate $y'$ at the point $(-3\sqrt{3},1)$

Substitute $x=-3\sqrt{3}$ and $y = 1$ into $y'$: $y'=-\left(\frac{1}{-3\sqrt{3}}\right)^{\frac{1}{3}}=\frac{1}{\sqrt{3}}$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(-3\sqrt{3},1)$ and $m=\frac{1}{\sqrt{3}}$. So $y - 1=\frac{1}{\sqrt{3}}(x + 3\sqrt{3})$, which simplifies to $y=\frac{1}{\sqrt{3}}x+4$.

Step5: Differentiate the second equation $x^{2}+3y^{2}=3$ implicitly for $y'$

Differentiate both sides with respect to $x$: $2x + 6yy'=0$. Solve for $y'$: $y'=-\frac{x}{3y}$.

Step6: Differentiate $y'$ to find $y''$

Using the quotient - rule $y''=\frac{-3y+3xy'}{9y^{2}}$.

Step7: Substitute $y'=-\frac{x}{3y}$ into the expression for $y''$

$y''=\frac{-3y + 3x\left(-\frac{x}{3y}\right)}{9y^{2}}=\frac{-3y^{2}-x^{2}}{9y^{3}}$. Since $x^{2}+3y^{2}=3$ (or $x^{2}=3 - 3y^{2}$), substitute $x^{2}$: $y''=\frac{-3y^{2}-(3 - 3y^{2})}{9y^{3}}=-\frac{1}{3y^{3}}$.

Answer:

The equation of the tangent line is $y=\frac{1}{\sqrt{3}}x + 4$ and $y''=-\frac{1}{3y^{3}}$.