The number of points of discontinuity of y = f(x) where $\mathrm{f}\left(\mathrm{x}\right)=\frac{3}{2{\mathrm{u}}^2+5\mathrm{u}-3}$  and $\mathrm{u}=\frac{1}{\mathrm{x}+2}$ is

u is discontinuous at x = - 2, f(x) is discontinuous

at $\mathrm{u}=\frac{1}{2},-3\text{ when }\mathrm{u}=-3\Rightarrow \mathrm{x}=-\frac{7}{3},\mathrm{u}=\frac{1}{2}\Rightarrow \mathrm{x}=0$

Hence f is discontinuous at $x=-\frac{7}{3},-2,0$