jmc

Moving all the terms to the left-hand side, we have $$\frac{1}{x(x+1)}-\frac{1}{(x+1)(x+2)}-\frac{1}{3}<0.$$To solve this inequality, we find a common denominator: $$\frac{3(x+2)-3x-x(x+1)(x+2)}{3x(x+1)(x+2)}<0,$$which simplifies to $$\frac{6-x(x+1)(x+2)}{3x(x+1)(x+2)}<0.$$To factor the numerator, we observe that $x=1$ makes the numerator zero, so $x-1$ is a factor of the expression. Performing polynomial division, we get $$6-x(x+1)(x+2)=-(x-1)(x^2+4x+6).$$Therefore, we want the values of $x$ such that $$\frac{(x-1)(x^2+4x+6)}{x(x+1)(x+2)}>0.$$Notice that $x^2+4x+6=(x+2{)}^2+2,$ which is always positive, so this inequality is equivalent to $$f(x)=\frac{x-1}{x(x+1)(x+2)}>0.$$To solve this inequality, we make the following sign table:\begin{array}{c|cccc|c} &$x$ &$x-1$ &$x+1$ &$x+2$ &$f(x)$ \\ \hline$x<-2$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+$\\ [.1cm]$-2<x<-1$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$-1<x<0$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$0<x<1$ &$+%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>1$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}Putting it all together, the solutions to the inequality are $$x\in \boxed{(-\infty ,-2)\cup (-1,0)\cup (1,\infty )}.$$