On
The derivative of $f(x)$ is
$$f'(x)=\frac 14+2x\sin\left(\frac 1x\right)-\cos\left(\frac 1x\right)$$
In any neighborhood of zero, you can find a very small positive $a$ where $\cos\left(\frac 1a\right)=1$. Then the second term of my expression for $f'(a)$ is near zero, thus $f'(a)$ is near $-\frac 34$, thus $f'(a)$ is negative, thus $f(x)$ is decreasing in a neighborhood of $a$ that is inside our neighborhood of zero.
In that same neighborhood of zero, you can find a very small positive $b$ where $\cos\left(\frac 1b\right)=-1$. Then the second term of my expression for $f'(b)$ is near zero, thus $f'(b)$ is near $\frac 54$, thus $f'(b)$ is positive, thus $f(x)$ is decreasing in a neighborhood of $b$ that is inside our neighborhood of zero.
Both $a$ and $b$ are positive, so $f(x)$ is continuous on the interval between $a$ and $b$. You can easily show that any function continuous on an interval that increases in a subinterval and also decreases in a subinterval is not injective on that interval. (Use the Intermediate Value Theorem on the four cases $a<b$ or $b<a$, $f(a)<f(b)$ or $f(b)<f(a)$). This is a useful theorem to use in many other problems, as well.
Thus your function $f(x)$ is not injective in any neighborhood of zero, in the positive part of that interval alone.
On
Take any interval $(-\epsilon, \epsilon )$. Then, since $f'(x)=\frac{1}{4}+2x\sin \frac{1}{x}-\cos \frac{1}{x}$, we may choose $x\in (0,\epsilon )\ $such that $\cos \frac{1}{x}=1$ and $\sin \frac{1}{x}=0\Rightarrow f'(x)<0$.
On the other hand, we may also choose $y\in (0,\epsilon)$ so that $\cos \frac{1}{y}=0\Rightarrow f'(y)>0$. Wlog $x<y$. Since $f$ is differentiable on $(x,y)$, it has a local minimum at some point $c\in (x,y)$ and now an application of the Intermediate Value Theorem to the intervals $(x,c)$ and $(c,y)$ gives the result.
When $-\frac 1n<x<\frac 1n$ the argument of $\sin (\frac 1x)$ takes all the values of the interval $]n,\infty[$ so that $\sin (\frac 1x)$ takes infinitely many times all the values of the rank $[-1,1]$ of the function $\sin x$ (If you want to draw the graph of $\sin (\frac 1x)$, you have a shaded rectangle in the neighborhood of $0$).
This is the reason for which the function $g(x)=x^2\sin (\frac 1x)$ can't be 1−1 for any neighborhood of $0$ and so is for $f(x)=\frac x4 +x^2\sin(\frac 1x)$.