If $x=\sqrt{6}+2$ and $y=\sqrt{6}-2$, then what is the value of ${(\frac{x}{y}+\frac{y}{x})}^2-3$?

To find the value of the expression ${(\frac{x}{y}+\frac{y}{x})}^2-3$, we are given:
$x=\sqrt{6}+2$
$y=\sqrt{6}-2$

Step 1: Simplify the inner expression $\frac{x}{y}+\frac{y}{x}$
Let us find a common denominator for the sum of the two fractions:
$\frac{x}{y}+\frac{y}{x}=\frac{x^2+y^2}{xy}$

Step 2: Calculate the product $xy$
Using the difference of squares identity, $(a+b)(a-b)=a^2-b^2$:
$xy=(\sqrt{6}+2)(\sqrt{6}-2)$
$xy={\left(\sqrt{6}\right)}^2-2^2$
$xy=6-4=2$

Step 3: Calculate the sum of squares $x^2+y^2$
We can find the individual squares first:
$x^2={(\sqrt{6}+2)}^2={\left(\sqrt{6}\right)}^2+2\left(2\right)\left(\sqrt{6}\right)+2^2=6+4\sqrt{6}+4=10+4\sqrt{6}$
$y^2={(\sqrt{6}-2)}^2={\left(\sqrt{6}\right)}^2-2\left(2\right)\left(\sqrt{6}\right)+2^2=6-4\sqrt{6}+4=10-4\sqrt{6}$
Now, add them together:
$x^2+y^2=(10+4\sqrt{6})+(10-4\sqrt{6})=20$

Step 4: Evaluate the value of $\frac{x}{y}+\frac{y}{x}$
Substitute the values we calculated into our step 1 fraction:
$\frac{x^2+y^2}{xy}=\frac{20}{2}=10$

Step 5: Compute the final expression value
Now, we substitute 10 back into the original expression:
${(\frac{x}{y}+\frac{y}{x})}^2-3={10}^2-3$
$=100-3$
$=97$

Therefore, the correct option is 97.