They occur at #x=\pm\frac{1}{\sqrt{3}}#. On the graph of #f#, they are at the points #(x,y)=(\pm\frac{1}{\sqrt{3}},f(\pm\frac{1}{\sqrt{3}}))=(\pm\frac{1}{\sqrt{3}},31/9)#.
The first derivative is #f'(x)=4x^{3}-4x# and the second derivative is #f''(x)=12x^{2}-4=4(3x^{2}-1)#. Setting #f''(x)=0# gives possible inflection points at #x=\pm\frac{1}{\sqrt{3}}#. That these are actual inflection points follows from the fact that #f''(x)# changes sign at these values of #x#. The corresponding #y#-coordinate of both of the inflection point is #f(\pm\frac{1}{\sqrt{3}})=31/9#.