1. Given function

$$f(x)=\frac{\log_{x+1}(x-2)}{e^{2\log_e x}-(2x+3)}, \quad x\in \mathbb R$$

We need the domain of this function.

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2. Condition for the logarithm

For $\log_{x+1}(x-2)$ to be defined, we need:

• Argument $x-2>0 \Rightarrow x>2$
• Base $x+1>0 \Rightarrow x>-1$
• Base $x+1\neq 1 \Rightarrow x\neq 0$

Combining these:

$x>2$

already ensures $x>-1$ and $x\neq 0$. So the logarithmic part requires

$x>2.$

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3. Condition for the exponential denominator

Denominator must be nonzero:

$e^{2\log_e x}-(2x+3)\neq 0$

Also, since $\log_e x$ appears, we must have

$x>0.$

But this is already satisfied by $x>2$.

Now simplify:

$e^{2\log_e x}=e^{\log_e x^2}=x^2$

More directly,

$2\log_e x=\log_e x^2 \quad (x>0),$

so

$e^{2\log_e x}=x^2.$

Thus denominator becomes

$x^2-(2x+3)=x^2-2x-3=(x-3)(x+1).$

For denominator nonzero:

$(x-3)(x+1)\neq 0 \Rightarrow x\neq 3,-1.$

But from $x>2$, only $x=3$ is relevant.

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4. Combine all conditions

From logarithm: $x>2$

From denominator: $x\neq 3$

Hence domain is

$(2,\infty)-\{3\}.$

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5. Check options

• A: $(-1,\infty)-\{3\}$ ❌ includes values between $-1$ and $2$ where $x-2\le 0$
• B: $\mathbb R-\{-1,3)$ ❌ invalid and does not match conditions
• C: $(2,\infty)-\{3\}$ ✅ correct
• D: $\mathbb R-\{3\}$ ❌ includes values where logarithm is undefined

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6. Final answer

$\boxed{(2,\infty)-\{3\}}$

So the correct option is C.