Let $$f$$ be a composite function of $$x$$ defined by $$f(u) = \frac{1}{u^2+u-2}$$, $$u(x) = \frac{1}{x-1}$$. Then the number of points $$x$$ where $$f$$ is discontinuous is :

$$u(x) = \frac{1}{x-1}$$

A rational function is discontinuous where its denominator is zero.

$$x - 1 = 0 \implies \mathbf{x = 1}$$

$$f(u) = \frac{1}{u^2+u-2}$$

$$f(u)$$ is discontinuous at $$u = -2$$ and $$u = 1$$.

Case $$u = -2$$: $$\frac{1}{x-1} = -2 \implies 1 = -2x + 2 \implies 2x = 1 \implies \mathbf{x = 0.5}$$

Case $$u = 1$$: $$\frac{1}{x-1} = 1 \implies 1 = x - 1 \implies \mathbf{x = 2}$$

Combining all identified points, the composite function is discontinuous at $$x \in \{0.5, 1, 2\}$$