Actually, you can't do
$$\sqrt{x} = x^\frac{1}{2} = x^\frac{2}{4} = \sqrt[4]{x^2}$$
in general for anything outside of the positive reals.
Rational Exponents
$x^x$ is defined somewhere on the real number line as long as $x$ is a negative integer or a rational number with an odd denominator when fully reduced. Any other value of $x$ will produce an answer that is undefined on the real number line.
To explain why, let me cut to the chase and bypass some technical math to help you visualize the intuition.
For starters, we need to look at the set of complex numbers. We can write complex numbers as $x+yi$, but they can also be represented in what is called polar form, as an angle and a magnitude.
Accept the following: When we multiply two complex numbers together, we multiply their magnitudes (not as important for this topic) and add their angles (very important). For example, the complex numbers at $(1, 30^\circ)$ and $(1, 40^\circ)$ will have a product located at $(1, 70^\circ)$.
Whenever you take a square root of a complex number, realize that there are two complex numbers you can square that will get you to that number. If the original complex number has an angle of $20^\circ$, the root could either be $10^\circ$ or $190^\circ$, because $190^\circ \cdot 2 = 380^\circ = 20^\circ$. You can see why the second angle solution is tricky, and so that's why we consider the $10^\circ$ solution the principal root, even though the other one can be squared to get the same value.
This is why the positive reals (with angle $0^\circ$) can have square roots of $0^\circ$ (the positive root), and $180^\circ$ (the negative root), and why we consider the positive root the principal one.
When you plot all of the roots of a complex number, they will evenly divide the $360^\circ$ circle around the root. For instance, here are where all the 5th roots of $-1$ are:
Here are the sixth roots of $-1$:
We can take any odd root and still hit $-1$, but we can't for an even root. Hence, the denominator must be odd in order for a real solution to exist.
$$
\frac{180^\circ + n \cdot 360^\circ}{2n+1} = \frac{(2n+1)180^\circ}{2n+1} = 180^\circ \\
\frac{180^\circ + k \cdot 360^\circ}{2n} = \frac{(2k+1)180^\circ}{2n} \ne 180^\circ
$$
The last statement cannot hold true for any $k$ value because $2k+1$ is odd and $2n$ is even, which means the expression cannot be a whole number.
This is also why your fraction must be simplified. Taking $-1$ to the $\frac{1}{2}$ power gives us angles of $90^\circ$ and $270^\circ$. Taking $-1$ to the $\frac{2}{4}$ power is multiplying the angle by $2$, then splitting into $4$, which gives us angles of $0^\circ$ and $180^\circ$ in addition to the other two, giving you extraneous solutions.
Real Exponents
Accept the following: In general, to convert a polar form complex number to rectangular form, we use the formula:
$$
re^{i\theta} = r(\cos\theta + i \sin\theta)
$$
Since $\theta$ can be any real value, this tells us that we can substitute values into this formula to obtain any value we want for a negative exponent, even an irrational one!
First note the following (we will use this later):
$$
-1 = e^{i \pi} \\
\ln(-1) = i \pi
$$
Technically, there are an infinite number of values for $\ln(-1)$, but we will once again take the simplest one, the principal solution.
Now use change of base (assume $x$ is negative):
$$
x^x = e^{x \ln x} \\
\ln x = \ln{(-x)} + \ln{(-1)} \\
\ln x = \ln{(-x)} + i\pi \\
e^{x \ln x} = e^{x\ln{(-x)} + i\pi x } = e^{x\ln{(-x)}} e^{ i\pi x } \\
$$
We can see that the $e^{x\ln{(-x)}}$ part is a real number. Let's look at the other part.
$$
e^{ i\pi x } = \cos{\pi x} + i \sin{\pi x}
$$
So finally, we can compute that for any negative $x$:
$$
x^x = e^{x\ln{(-x)}}(\cos{\pi x} + i \sin{\pi x})
$$
This is now a well-defined complex number!
Calculator Issues?
The reason why your calculator has breaks in it, is because of how it tests and samples values. If it picks decimal values, it will never be able to evaluate any of these on the real number line except for the negative integer values. If it can somehow do these complex operations and can test exact rational values by the formula I gave using symbolic evaluation, then it will give you some values for the ones it tests. But otherwise, you won't have much luck, since most all of these values will give you complex numbers!
Best Answer
I feel the need to be pedantic here, and want to make a distinction between the expression $$\frac{(x^2-121)/(x^2-4)}{(x+2)/(x-11)}$$ and the rational function it represents.
The expression is undefined at $x=11$, since we cannot evaluate $(x+2)/(x-11)$ when $x=11$.
However the rational function it represents can also be written as $$\frac{(x^2-121)(x-11)}{(x+2)(x^2-4)},$$ so the rational function is defined at $x=11$, because it can be written as a fraction of polynomials for which $11$ is not a root of the polynomial in the denominator.
There are several good reasons to make this distinction, though I'm not sure it's usually made at a high school level, but I'll cite Wikipedia as a source to show that this distinction is in fact made.
An aside on why this distinction is made
The short version is that algebraically, we can think of rational functions as fractions of polynomials. Thus if two fractions are equal, then the functions they define should also be equal.
I.e., because these fractions of polynomials are equal, $$\frac{1}{x-2} = \frac{x+2}{x^2-4}$$ they should represent the same function.
This is related to the idea that polynomials aren't functions. See this question for more on that topic.