Let $$f : R \to R$$ be defined as $$f(x) = \begin{cases} \frac{x^3}{(1-\cos 2x)^2} \log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right), & x \neq 0 \\ \alpha, & x = 0 \end{cases}$$
If $$f$$ is continuous at $$x = 0$$, then $$\alpha$$ is equal to:

For continuity at $$x=0,$$ we require

$$\alpha=\lim_{x\to0}\frac{x^3}{(1-\cos2x)^2}\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$$

Using $$1-\cos2x=2\sin^2x,$$

we get $$(1-\cos2x)^2=4\sin^4x$$

Hence,

$$\alpha=\lim_{x\to0}\frac{x^3}{4\sin^4x}\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$$

Therefore,

$$\alpha=\frac14\lim_{x\to0}\left(\frac{x}{\sin x}\right)^4\frac1x\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$$

Since

$$\lim_{x\to0}\frac{\sin x}{x}=1,$$

we obtain

$$\alpha=\frac14\lim_{x\to0}\frac1x\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)$$

Using logarithmic properties,

$$\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)=\log_e(1+2xe^{-2x})-2\log_e(1-xe^{-x})$$

Now,

$$e^{-2x}=1-2x+O(x^2),\qquad e^{-x}=1-x+O(x^2)$$

Hence,

$$2xe^{-2x}=2x+O(x^2),\qquad xe^{-x}=x+O(x^2)$$

Using

$$\log(1+t)=t+O(t^2),$$

we get

$$\log_e(1+2xe^{-2x})=2x+O(x^2)$$

and

$$-2\log_e(1-xe^{-x})=2x+O(x^2)$$

Therefore,

$$\log_e\left(\frac{1+2xe^{-2x}}{(1-xe^{-x})^2}\right)=4x+O(x^2)$$

Thus,

$$\alpha=\frac14\lim_{x\to0}\frac{4x+O(x^2)}x$$

$$=\frac14\cdot4$$

$$=1$$