The problem requires solving an exponential equation for the variable $x$ and then evaluating a specific expression involving $x$. We need to find the value of $\frac{2^{x^2}}{(2^{x})^2}$.

Solving Exponential Equation $3^{x^2} = 27 \times 9^x$

First, rewrite the equation with a common base, which is 3:

• $27 = 3^3$
• $9 = 3^2$

Substitute these into the equation:

$3^{x^2} = 3^3 \times (3^2)^x$

Using the exponent rule $(a^m)^n = a^{m \times n}$:

$3^{x^2} = 3^3 \times 3^{2x}$

Using the exponent rule $a^m \times a^n = a^{m+n}$:

$3^{x^2} = 3^{3+2x}$

Since the bases are equal, the exponents must be equal:

$x^2 = 3 + 2x$

Rearrange this into a standard quadratic equation:

$x^2 - 2x - 3 = 0$

Factor the quadratic equation:

$(x-3)(x+1) = 0$

The possible values for $x$ are $x = 3$ or $x = -1$.

Evaluating Expression $\frac{2^{x^2}}{(2^{x})^2}$

Simplify the expression using exponent rules:

First, simplify the denominator: $(2^x)^2 = 2^{2x}$.

The expression becomes:

$\frac{2^{x^2}}{2^{2x}}$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$2^{x^2 - 2x}$

From the quadratic equation step, we found that $x^2 - 2x = 3$. Substitute this value into the expression:

$2^3$

The value of the expression is $2^3$, which is equal to 8.