= Solution
Near zero, $x^3-x^5=x^3(1-x^2)$ has the sign of $x$, so $f(x)<1$ just to the left and $f(x)>1$ just to the right. Thus the <critical point> is neither a <local minimum> nor a <local maximum>. Moreover the <Taylor series> gives $f''(x)=6x+O(x^3)$, which changes sign at zero. Therefore
$$
\boxed{(0,1)\text{ is a stationary inflection point}.}
$$
This is the point called a “saddle point” in the question; <stationary inflection point> specifies the one-variable meaning.
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