Why have you not shown any work at all? Did you not even try this? I presume, if you were given this problem in a class, that you know that "stationary points" are points where the gradient is 0:
\(\displaystyle f(x, y, z)= x^2y+ y^2z+ z^2x\)
\(\displaystyle f_x= 2xy+ z^2= 0\)
\(\displaystyle f_y= x^2+ 2yz= 0\)
\(\displaystyle f_z= y^2+ 2zx= 0\)
From the first equation \(\displaystyle y= -\frac{z^2}{2x}\). Replacing y with that in the second equation \(\displaystyle x^2- \frac{z^3}{x}= 0\) so \(\displaystyle z^3= x^3\). x, y, and z are real so \(\displaystyle z= x\). Then \(\displaystyle y= -\frac{z^2}{2x}= -\frac{x^2}{2x}= -\frac{x}{2}\) and the third equation is \(\displaystyle \frac{x^2}{4}+ 2x^2= \frac{9}{4}x^2= 0\). x= y= z= 0 is the only stationary point. On the line x= y= z, \(\displaystyle f(x, x, x)= 3x^3\) which is positive for x positive and negative for x negative. The stationary point is a saddle point.
