Answer
There is a point on inflection at $x$=$0$.
Work Step by Step
When $y$=$x^3-3x+3$,
then $y'$=$3x^2-3$=$3(x-1)(x+1)$ and
$y''$=$ 6x $.
The curve rises on $(-\infty,-1)U(1,\infty)$ and falls on $(-1,1)$.
At $x$=$-1$ there is
a local maximum and at $x$=$1$ a local minimum.
The curve is concave down on $(-\infty ,0)$ and concave up on
$(0,\infty)$.
There is a point on inflection at $x$= $0$.
