 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 $f^{\prime \prime }(x)=6x+O(x^3)$, which changes sign at zero. ThereforeThis is the point called a “saddle point” in the question; stationary inflection point specifies the one-variable meaning.