1. Interpret the function carefully

The given function is

$$f(x)= \begin{cases} xe^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)}, & x\neq 0 \\ 0, & x=0 \end{cases}$$

We must check continuity and differentiability, especially at $x=0$.

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2. Simplify the expression for $x>0$ and $x<0$

Since $|x|$ behaves differently on the two sides of $0$:

• If $x>0$, then $|x|=x$, so

  $$\frac{1}{|x|}+\frac{1}{x}=\frac{1}{x}+\frac{1}{x}=\frac{2}{x}.$$

  Hence,

  $$f(x)=xe^{-2/x}, \quad x>0.$$

• If $x<0$, then $|x|=-x$, so

  $$\frac{1}{|x|}=\frac{1}{-x}=-\frac{1}{x}.$$

  Therefore,

  $$\frac{1}{|x|}+\frac{1}{x}=-\frac{1}{x}+\frac{1}{x}=0.$$

  Hence,

  $$f(x)=xe^0=x, \quad x<0.$$

So the function becomes

$$f(x)= \begin{cases} x, & x<0,\\ 0, & x=0,\\ xe^{-2/x}, & x>0. \end{cases}$$

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3. Check continuity at $x=0$

We compute left-hand and right-hand limits.

Left-hand limit

For $x<0$, $f(x)=x$. Thus

$$\lim_{x\to 0^-} f(x)=\lim_{x\to 0^-} x=0.$$

Right-hand limit

For $x>0$, $f(x)=xe^{-2/x}$. As $x\to 0^+$,

$$e^{-2/x}\to 0$$

very rapidly, so

$$\lim_{x\to 0^+} xe^{-2/x}=0.$$

Also,

$$f(0)=0.$$

Thus,

$$\lim_{x\to 0} f(x)=f(0)=0.$$

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

For all $x\neq 0$, the formulas $x$ and $xe^{-2/x}$ are standard continuous functions. Hence $f$ is continuous for all real $x$.

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4. Check differentiability at $x=0$

Use the definition:

$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h} = \lim_{h\to 0}\frac{f(h)}{h}.$$

Now evaluate from both sides.

Left-hand derivative

For $h<0$, $f(h)=h$. Hence

$$\frac{f(h)}{h}=\frac{h}{h}=1.$$

So,

$$\lim_{h\to 0^-}\frac{f(h)-f(0)}{h}=1.$$

Right-hand derivative

For $h>0$, $f(h)=he^{-2/h}$. Hence

$$\frac{f(h)}{h}=e^{-2/h}.$$

So,

$$\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+} e^{-2/h}=0.$$

Since left-hand derivative $\neq$ right-hand derivative,

$$f'(0) \text{ does not exist.}$$

So $f$ is not differentiable at $x=0$.

For $x\neq 0$, it is differentiable because each branch is differentiable there.

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5. Conclusion

The function is:

• continuous for all $x$,
• not differentiable at $x=0$.

Therefore, the correct option is

$$\boxed{\text{C}}$$

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6. Compare with stored correct answer

Stored correct answer: $\text{C}$

Our derived answer: $\text{C}$

They agree.