Example 1 Let $x, y, z \in \mathbf{R}^{+}, \sqrt{x^{2}+y^{2}}+z=1$, try to find the maximum value of $x y+2 x z$.


Let $z=\sin ^{2} \alpha, \sqrt{x^{2}+y^{2}}=\cos ^{2} \alpha, x=\cos ^{2} \alpha \sin \beta, y=\cos ^{2} \alpha \cos \beta, \alpha, \beta \in\left(0, \frac{\pi}{2}\right]$,
then $x y+2 x z=x(y+2 z)=\cos ^{2} \alpha \sin \beta\left(\cos ^{2} \alpha \cos \beta+2 \sin ^{2} \alpha\right)$
$$
\begin{array}{l}
=\frac{\sin \beta}{2-\cos \beta}(2-\cos \beta) \cos ^{2} \alpha\left(\cos \beta \cos ^{2} \alpha+2 \sin ^{2} \alpha\right) \\
=\frac{\sin \beta}{2-\cos \beta}\left(2 \cos ^{2} \alpha-\cos \beta \cos ^{2} \alpha\right)\left(\cos \beta \cos ^{2} \alpha+2 \sin ^{2} \alpha\right) \\
\leqslant \frac{\sin \beta}{2-\cos \beta}\left(\frac{2 \cos ^{2} \alpha-\cos \beta \cos ^{2} \alpha+\cos \beta \cos ^{2} \alpha+2 \sin ^{2} \alpha}{2}\right)^{2}=\frac{\sin \beta}{2-\cos \beta}
\end{array}
$$

Constructing the slope, we get that when $x=\frac{\sqrt{3}}{3}, y=z=\frac{1}{3}$, $x y+2 x z$ reaches its maximum value $\frac{\sqrt{3}}{3}$.