1. Given piecewise function

We need $f(x)$ to be differentiable at every point. Since each piece is differentiable in its own interval, we only need to check the junction points:

$x=1 \quad \text{and} \quad x=-1.$

2. Continuity at $x=1$

For differentiability, continuity is necessary.

• From the outer piece: $f(1)=\frac{1}{|1|}=1$
• From the inner piece as $x\to 1^-$: $a(1)^2+b=a+b$

So continuity at $x=1$ gives

$a+b=1 \qquad (1)$

3. Differentiability at $x=1$

For $x>1$, since $|x|=x$,

$f(x)=\frac{1}{x}$

Hence right-hand derivative at $x=1$ is

$f'_+(1)=\left(-\frac{1}{x^2}\right)_{x=1}=-1.$

For $|x|<1$,

$f(x)=ax^2+b \implies f'(x)=2ax.$

So left-hand derivative at $x=1$ is

$f'_-(1)=2a.$

Differentiability at $x=1$ requires

$2a=-1 \implies a=-\frac{1}{2}. \qquad (2)$

Using (1):

$-\frac{1}{2}+b=1 \implies b=\frac{3}{2}.$

4. Check at $x=-1$

Now verify continuity and differentiability there.

Continuity at $x=-1$

Outer piece gives

$f(-1)=\frac{1}{|-1|}=1.$

Inner piece gives

$a(-1)^2+b=a+b=-\frac{1}{2}+\frac{3}{2}=1.$

So continuous at $x=-1$.

Differentiability at $x=-1$

For $x<-1$, we have $|x|=-x$, so

$f(x)=\frac{1}{|x|}=\frac{1}{-x}=-\frac{1}{x}.$

Thus

$f'(x)=\frac{1}{x^2}$ for $x<-1$, hence left-hand derivative at $x=-1$ is

$f'_-( -1)=\frac{1}{(-1)^2}=1.$

For the inner piece,

$f'(x)=2ax.$

So right-hand derivative at $x=-1$ is

$f'_+( -1)=2a(-1)=-2a=-2\left(-\frac{1}{2}\right)=1.$

Thus derivatives match at $x=-1$ as well.

5. Final values

$a=-\frac{1}{2}, \qquad b=\frac{3}{2}.$

So the correct option is

$\boxed{\text{D}}.$