$\begin{array}{l}\text{If}y=\frac{e^{-x}+e^x}{e^{-x}-e^x}\text{, then}\frac{dy}{dx}\text{is equal to}\end{array}$

The correct option is:
$$\frac{4}{{(e^{-x}-e^x)}^2}$$

Step-by-Step Explanation:

We are given the function:
$$y=\frac{e^{-x}+e^x}{e^{-x}-e^x}$$

To find the derivative $\frac{dy}{dx}$, we apply the quotient rule of differentiation. The quotient rule states that for a function of the form $y=\frac{u}{v}$:
$$\frac{dy}{dx}=\frac{u^{\text{'}}v-u{v}^{\text{'}}}{v^2}$$

Let us define the numerator and denominator functions:
$u=e^{-x}+e^x$
$v=e^{-x}-e^x$

Now, we differentiate $u$ and $v$ with respect to $x$:
$$u^{\text{'}}=\frac{d}{dx}(e^{-x}+e^x)=-e^{-x}+e^x$$
$$v^{\text{'}}=\frac{d}{dx}(e^{-x}-e^x)=-e^{-x}-e^x=-(e^{-x}+e^x)$$

Next, we substitute these components into the quotient rule formula:
$$\frac{dy}{dx}=\frac{(-e^{-x}+e^x)(e^{-x}-e^x)-(e^{-x}+e^x)[-(e^{-x}+e^x\left)\right]}{{(e^{-x}-e^x)}^2}$$

Simplify the terms in the numerator:
Notice that $-e^{-x}+e^x=-(e^{-x}-e^x)$.
So, the first part of the numerator becomes:
$$-{(e^{-x}-e^x)}^2$$

The second part of the numerator becomes:
$$-[-{(e^{-x}+e^x)}^2]={(e^{-x}+e^x)}^2$$

Combining these, the numerator simplifies to:
$${(e^{-x}+e^x)}^2-{(e^{-x}-e^x)}^2$$

We can expand both squares using the identities ${(a+b)}^2=a^2+2ab+b^2$ and ${(a-b)}^2=a^2-2ab+b^2$:
$${(e^{-x}+e^x)}^2=e^{-2x}+2\left(e^{-x}\right)\left(e^x\right)+e^{2x}=e^{-2x}+2+e^{2x}$$
$${(e^{-x}-e^x)}^2=e^{-2x}-2\left(e^{-x}\right)\left(e^x\right)+e^{2x}=e^{-2x}-2+e^{2x}$$

Subtracting the two expanded expressions:
$$(e^{-2x}+2+e^{2x})-(e^{-2x}-2+e^{2x})=4$$

Substituting this simplified numerator back into the quotient formula yields:
$$\frac{dy}{dx}=\frac{4}{{(e^{-x}-e^x)}^2}$$