1. Given function

We have

We must check which option is true.

2. Check differentiability at $x=0$

To test differentiability at $0$,

Now

So

Since $|h\sin(1/h)|\le |h|\to 0$, we get

Hence $f'(0)=0$.

So $f$ is differentiable at $x=0$.

Therefore, Option A is false.

3. Find derivative for $x\ne 0$

For $x\ne 0$,

Differentiate:

Using product rule,

Thus

So

As $x\to 0$, the term $x(4+2\sin(1/x))\to 0$, so the sign of $f'(x)$ near $0$ is governed mainly by $\cos(1/x)$, which oscillates.

4. Check Option B: Is $f$ decreasing on some $(0,\delta)$?

For $f$ to be decreasing on $(0,\delta)$, we would need $f'(x)\le 0$ for all sufficiently small positive $x$.

But $\cos(1/x)$ oscillates between $-1$ and $1$ infinitely often near $x=0$.

Take sequences approaching $0^+$:

• If

then

and hence

for all sufficiently large $n$.

So arbitrarily close to $0^+$, $f'(x)$ is positive. Therefore $f$ cannot be decreasing on any interval $(0,\delta)$.

Thus Option B is false.

5. Check Option C: For any $\delta>0$, is $f$ not increasing on $(-\delta,0)$?

If $f$ were increasing on $(-\delta,0)$, we would need $f'(x)\ge 0$ throughout that interval.

Again use oscillation. Take a sequence approaching $0^-$ such that

with $x_n<0$ for suitable negative values, equivalently choose

Then

so

This only shows positive derivative somewhere, not enough to disprove increasing.

To disprove increasing, we need points with negative derivative too. Choose

Then

so

For sufficiently large $n$, $|y_n|$ is very small, hence

Thus, in every interval $(-\delta,0)$, there are points where $f'(x)<0$. So $f$ cannot be increasing on any such interval.

Hence Option C is true.

6. Check Option D: Is $x=0$ a local minimum?

We compare $f(x)$ with $f(0)=2$ near $0$. For $x\ne 0$,

Since

we have

Therefore for every $x\ne 0$,

So nearby values are always less than $f(0)$. Hence $x=0$ is actually a local maximum, not a local minimum.

Thus Option D is false.

7. Conclusion

The only true statement is $\boxed{\text{C}}.$