Solution

First and foremost it is important to determine the domain of the function:

\[ \solve{ -\displaystyle\frac{x^2}{{9}}+1&\geq& 0\\ -\displaystyle\frac{x^2}{{9}}&\geq&-1\\ \displaystyle\frac{x^2}{{9}}&\leq&1\\ x^2&\leq&9\\ x\leq3 &&x\geq -3 } \]

Since the domain is \([-3,3]\), we should only select values in that interval. I choose to use very specific values to get "nice" answers, which is why there are some inputs with radicals.

\[ \begin{array}{{c|c|c|c}} x&y=&... & y\\ \hline 0 & 4\sqrt{-\displaystyle \frac{(0)^2}{{9}} +1}& 4\sqrt{{1}} &4 \\ -3 & 4\sqrt{-\displaystyle \frac{(-3)^2}{{9}} +1}&4\sqrt{{0}}&0\\ 3&4\sqrt{-\displaystyle \frac{(3)^2}{{9}} +1}&4\sqrt{{0}}&0 \\ -\sqrt{{5}} &4\sqrt{-\displaystyle \frac{(-\sqrt{{5}})^2}{{9}}+1} &4\sqrt{\frac{{4}}{{9}}}&\frac{{8}}{{3}} \\ \sqrt{{5}} &4\sqrt{-\displaystyle \frac{(\sqrt{{5}})^2}{{9}}+1} &4\sqrt{\frac{{4}}{{9}}}&\frac{{8}}{{3}} \\-\sqrt{{8}} &4\sqrt{-\displaystyle\frac{(-\sqrt{{8}})^2}{{9}}+1}&4\sqrt{\frac{{1}}{{9}}}&\frac{{4}}{{3}}\\ \sqrt{{8}} &4\sqrt{-\displaystyle\frac{(\sqrt{{8}})^2}{{9}}+1}&4\sqrt{\frac{{1}}{{9}}}&\frac{{4}}{{3}} \end{array} \]

With these values, we can create the following sketch:

Once we have the basic shape, we can fill it in: