Example 18 Let positive real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$.
Let $f(x)=\frac{x}{1-x^{2}}, x \in(0,1)$, then $f^{\prime}(x)=\frac{1+x^{2}}{\left(1-x^{2}\right)^{2}}, f^{\prime \prime}(x)=\frac{2 x}{\left(1-x^{2}\right)^{2}}+$ $\frac{4 x\left(1+x^{2}\right)}{\left(1-x^{2}\right)^{2}}$.

Although $f^{\prime}(x)>0$, we know that $f(x)$ is a convex function in $(0,1)$, but it is not easy to apply Jensen's inequality for the condition $x^{2}+y^{2}+z^{2}=1$. Therefore, we consider another approach, and look at the function $g(x)=\frac{1}{1-x^{2}}, x \in(0,1)$, then $g^{\prime}(x)=\frac{2 x}{\left(1-x^{2}\right)^{2}}$.
Notice that the tangent line of the function $g(x)$ at $x=\frac{\sqrt{3}}{3}$ is $y=g^{\prime}\left(\frac{\sqrt{3}}{3}\right)\left(x-\frac{\sqrt{3}}{3}\right)+g\left(\frac{\sqrt{3}}{3}\right)=$ $\frac{3 \sqrt{3}}{2} x$.
By $\frac{1}{1-x^{2}}-\frac{3 \sqrt{3}}{2} x=\frac{2-3 \sqrt{3} x\left(1-x^{2}\right)}{2\left(1-x^{2}\right)}$ and let $h(x)=3 \sqrt{3} x\left(1-x^{2}\right)$, and when $0<x<\frac{\sqrt{3}}{3}$, $h^{\prime}(x)>0$, when $\frac{\sqrt{3}}{3}<x<1$, $h^{\prime}(x)<0$, we know that $h(x)$ reaches its maximum value 2 at $x=\frac{\sqrt{3}}{3}$, i.e., $2-3 \sqrt{3} x\left(1-x^{2}\right) \geqslant 0$, hence $\frac{1}{1-x^{2}} \geqslant \frac{3 \sqrt{3}}{2} x$, which also means $\frac{x}{1-x^{2}} \geqslant \frac{3 \sqrt{3}}{2} x^{2}$.
Thus, $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2}\left(x^{2}+y^{2}+z^{2}\right)=\frac{3 \sqrt{3}}{2}$.
Therefore, when $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$ reaches its minimum value $\frac{3 \sqrt{3}}{2}$.