We need to find the points where $f(x)=|x^2-2x-3| \, e^{|9x^2-12x+4|}$ is not differentiable.

1. Simplify the expression

First factor the terms inside modulus.

First modulus

$x^2-2x-3=(x-3)(x+1)$ So, $|x^2-2x-3|=|(x-3)(x+1)|$ This can cause non-differentiability where $x^2-2x-3=0 \implies x=3,-1.$

Second modulus

$9x^2-12x+4=(3x-2)^2$ Hence $|9x^2-12x+4|=|(3x-2)^2|=(3x-2)^2$ because a square is always non-negative.

So the function becomes $f(x)=|x^2-2x-3| \, e^{(3x-2)^2}.$

2. Identify possible non-differentiable points

Now,

• $e^{(3x-2)^2}$ is differentiable for all real $x$.
• Therefore, any non-differentiability can only come from $|x^2-2x-3|.$

Let $g(x)=x^2-2x-3.$ Then $f(x)=|g(x)|\,e^{(3x-2)^2}.$

We know that $|g(x)|$ is not differentiable at points where $g(x)=0 \quad \text{and} \quad g'(x)\neq 0.$

The zeros are: $g(x)=0 \implies x=-1,3.$

Also, $g'(x)=2x-2.$

Check at these points:

• At $x=-1$: $g'(-1)=-4\neq 0$ so $|g(x)|$ is not differentiable at $x=-1$.

• At $x=3$: $g'(3)=4\neq 0$ so $|g(x)|$ is not differentiable at $x=3$.

Thus $f(x)$ is not differentiable at exactly these two points, since multiplying by the smooth nonzero function $e^{(3x-2)^2}$ does not remove the cusp.

3. Count the number of such points

The function is not differentiable at $x=-1,\;3.$

So the number of points is $2.$

4. Match with options

• A: four points
• B: three points
• C: two points
• D: one point

Therefore, the correct option is: $\boxed{\text{C: two points}}$

5. Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.