There is the little preliminary question what $x^{2/{\rm odd number}}$ means when $x$ is negative. Putting this aside we can restrict to $\bigl({\mathbb R}_{\geq0}\bigr)^2$.
So we are told to "observe" the three curves $x^2+y^2=a^2\ ,\quad x^{2/3}+y^{2/3}=a^{2/3}\ ,\quad x^{2/5}+y^{2/5}=a^{2/5}\qquad(*)$ in the first quadrant. Note that we can solve for $y$ in the form $y=\bigl(a^{2/n}-x^{2/n}\bigr)^{n/2}\qquad(0\leq x\leq a)$ in each of the cases $n=1,\ 3,\ 5$, so that each of these curves appears as a graph. Looking at the equations $(*)$ we see that all three graphs will be symmetric with respect to the line $x=y$ and connect the point $(0,a)$ monotonically decreasing with $(a,0)$. See the following figure.

The case $n=1$ is obvious: The curve is a quarter circle. Its tangent at $(0,a)$ is horizontal and at $(a,0)$ vertical. This is different from the cases $n=3$ and $n=5$: For $f_n(x):=\bigl(a^{2/n}-x^{2/n}\bigr)^{n/2}$ we have $f_n'(x)=-{2\over n}\bigl(a^{2/n}-x^{2/n}\bigr)^{{n\over2}-1}\ x^{{2\over n}-1}\qquad (0 and therefore $f_n'(a)=0$ when $n>2$. It follows that in these cases the tangent is horizontal at $(a,0)$ and, by symmetry, vertical at $(0,a)$.