jmc

We can factor as follows: $$\begin{array}{rl}f(x,y,z)& =x^{2}y+y^{2}z+z^{2}x-x{y}^2-y{z}^2-z{x}^2\\ & =x^{2}y-x{y}^2+y^{2}z-z{x}^2+z^{2}x-y{z}^2\\ & =xy(x-y)+z(y^2-x^2)+z^2(x-y)\\ & =xy(x-y)-z(x-y)(x+y)+z^2(x-y)\\ & =(x-y)(xy-xz-yz+z^2)\\ & =(x-y)(x-z)(y-z).\end{array}$$The expression has cyclic symmetry (meaning that if we replace $(x,y,z)$ with $(y,z,x)$, then it remains the same), so we can assume that $x\ge y$ and $x\ge z.$ Thus, $x-y\ge$ and $x-z\ge 0.$

If $y<z,$ then $f(x,y,z)\le 0,$ so assume that $y\ge z.$ Then by AM-GM, $$(x-y)(y-z)\le {\left(\frac{(x-y)+(y-z)}{2}\right)}^2=\frac{(x-z{)}^2}{4},$$so $$(x-y)(x-z)(y-z)\le \frac{(x-z{)}^3}{4}\le \frac{1}{4}.$$Equality occurs when $x=1,$ $y=\frac{1}{2},$ and $z=0,$ so the maximum value is $\boxed{\frac{1}{4}}.$