jmc

We notice that the numerator and denominator share common factors: $x^2+x^3-2x^4=x^2(1+x-2x^2)$ and $x+x^2-2x^3=x(1+x-2x^2).$ Hence, whenever $x(1+x-2x^2)\ne 0,$ we can write $$\frac{x^2+x^3-2x^4}{x+x^2-2x^3}=\frac{x^2(1+x-2x^2)}{x(1+x-2x^2)}=x.$$It follows that the given inequality is satisfied if and only if $x\ge -1$ and $x(1+x-2x^2)\ne 0.$ The roots of $1+x-2x^2$ are $x=1$ and $x=-\frac{1}{2},$ so we cannot have $x=0,$ $x=1,$ or $x=-\frac{1}{2}.$ Putting all this together, the solution set of the inequality consists of the interval $[-1,\infty )$ with three "holes": $$x\in \boxed{[-1,-\frac{1}{2})\cup (-\frac{1}{2},0)\cup (0,1)\cup (1,\infty )}.$$