= Solution
A <stationary point> is a point where the function is differentiable and its first <derivative> vanishes. Interpret the cube root as the real <cube root>, and put $q(x)=x^2-8x+7$. Then $f(x)=15+(x-3)\sqrt[3]{q(x)}$. For $x\ne1,7$, the <product rule> and <chain rule> give
$$
f'(x)=\frac{3q(x)+(x-3)q'(x)}{3\big(\sqrt[3]{q(x)}\big)^2}
=\frac{5x^2-38x+45}{3\big(\sqrt[3]{q(x)}\big)^2}.
$$
The denominator is positive wherever it is defined. Consequently the two stationary abscissae are
$$
\boxed{x_\pm=\frac{19\pm2\sqrt{34}}5.}
$$
Using the stationary equation to simplify the value of $q$ gives $q(x_\pm)=-2(x_\pm+5)/5$. Thus the complete stationary points are
$$
\boxed{\left(x_\pm,\ 15-(x_\pm-3)\sqrt[3]{\frac{2(x_\pm+5)}5}\right).}
$$
Numerically they are approximately $(1.46762,17.10362)$ and $(6.13238,9.84664)$. The numerator of $f'$ is positive outside $[x_-,x_+]$ and negative inside, so the first is a local maximum and the second a local minimum.
At $x=1,7$, the real function is continuous but has a vertical tangent; the factor $x-3$ does not cancel either cube-root singularity. Those points are not stationary points. This is the distinction required when finding <stationary points with real cube-root factors>.
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