How to use differentiation to show that a curve is symmetric above x axis

Question

If $x^2+y^2=9$, how can I show that this curve is symmetric above $x$-axis using differentiation.

Am I correct?

Clearly $\frac{dy}{dx} =-\frac{x}{y}$ . So for all $x$ in the domain, $y=+$ or $-$. So I have two same values with only a sign difference. Can I conclude by this reasoning that the curve is symmetric above $x$?

Or is there a good reasoning way to use differentiation to prove it.

Answer 1

Consider the curves given by $$\begin{array} \\ x^2+y^2-9=0 & x\ge0 & y >0\\ x^2+(y-3)^2-9=0 & x\le0 & y>3 \end{array}$$

Here is the plot. So as you can see that the slope of tangents are equal at all values of $x=\pm a$. But they aren't symmetric.

So showing slopes are equal isn't a guarantee for symmetry. You must show that the values $f(a)=f(-a)$ to show symmetry.

Here is another example.