First, you've done something great: you've looked at a problem that probably wasn't in your textbook and analyzed it to death and made mistakes and corrected them. That's terrific!
OK, on to answer your questions (and to counter-answer a few).
First, the notion of "tangent" that you're using isn't generally called (by folks doing calculus) "tangent". Your idea of a line $L$ be tangent to a curve $C$ is that $L$ and $C$ share a point $P$ and at the point $P$ they have the same slope and near $P$, the curve $C$ is on one side of $L$. That's almost right...except that the "one side of" clause isn't included in the definition. So we often say that if $f$ is a function whose graph is a curve $C$, and $L$ and $C$ meet at $P$ and have the same slope at $P$, then "$L$ is the tangent to $f$ at $P$", whether $L$ and $C$ cross there or not.
Still, for polynomials it's an interesting and useful concept.
Let me switch to talking about functions.
If $f(x)$ is a polynomial function and $g(x) = mx + b$ is a line (i.e., $m$ is nonzero), and $f(a) = g(a)$ for some $a$, then we say that $f$ and $g$ intersect, or "agree to order zero"; if $f(a) = g(a) = 0$, we say that they have a shared root. If the slopes of $f$ and $g$ are the same, then we say that they "agree to first order." In this case, the equation $f(x) - g(x) = 0$ has (at least) a double-root at $a$, and indeed, for polynomials you could call this a "definition" of the derivative: the linear function $g$ with the property that $f(x) - g(x)$ has a double root at $a$ [or the slope of that line] could be called the derivative of $f$ at $a$.
The root could have higher order; it could be a triple root, or a fourth root, etc. And as you have observed, for odd-count repeated roots (when $f$ is polynomial), this indicates a crossing at $x = a$, while for even-count repeated roots, it indicates that near $x = a$, the graph of $f$ lies on one side of its tangent. By the time you've studied the second derivative and the Mean Value Theorem, you should be able to prove this.
What about those extra roots? Well, if you're willing to consider complex numbers, things actually get easier. Every degree-$n$ complex polynomial has $n$ roots, at least when you count repeated roots the appropriate number of times. So if $f$ is a cubic, when you look at
$$
f - g
$$
you've still got a cubic. And if they meet and have the same slope at $x = a$, then you get a double-root there...so there has to be a third root somewhere else. There's one exception: look at $f(x) = x^3$ and $g(x) = 0$. These meet at $a = 0$, and $f - g$ has a triple root there. So the "extra" root has gotten lumped in with the double-root at the tangency.
Since every real root must also be a complex root, every real degree $n$ polynomial has at most $n$ roots. So if you have a line tangent to the graph of such a polynomial, there are at most $n-2$ "other meetings" between that tangent line and the graph of the polynomial.
To go in a different direction, not every function is a polynomial. There are functions that touch their tangent lines in very odd ways. Consider the function
$$
f(x) = \begin{cases} 0 & x = 0 \\ x^2 \sin \frac{1}{x^2} & x \ne 0 \end{cases}
$$
Its tangent at the point $x = 0$ is the $x$-axis (which takes some work to show!), but this tangent intersect the graph of $f$ infinitely often in any neighborhood of $x = 0$. Yikes!
There is also a standard example (which I won't write down, because it probably uses things you haven't yet encountered) of a nonconstant function $f$ with the property that $f(0) = 0$ and $f'(0) = 0$, so its tangent line at $0$ is the $x$-axis. But at $x = 0$, the graph of $f$ meets its tangent "to infinite order," i.e., if you approximate $f$ near zero as well as possible with a degree $n$ polynomial, that polynomial will have $n$ roots all at $0$...and this is true for any $n$. We say that the function $f$ is "extremely flat" at $x = 0$. Such functions are actually important for "constructing partitions of unity" (whatever that might mean!) later on, i.e., they're not just wacko examples, but actually represent important phenomena in mathematics.
My answer's now almost as long as your question, and I hope I've addressed most of what you asked.
My personal favorite book for helping folks understand questions like this is Calculus, by Michael Spivak. It might well be better called something like "An introduction to analysis", but whatever you call it, it's a treasure.
Best Answer
$$(x+y+z)^2=9=x^2+y^2+z^2+2(xy+yz+zx)\implies xy+yz+zx=3$$ $$(x+y+z)^3=27=6xyz+3(x+y+z)(x^2+y^2+z^2)-2(x^3+y^3+z^3)=6xyz+27-6\implies xyz=1$$ Hence by Viète's formulas $x,y,z$ are the roots of $t^3-3t^2+3t-1=(t-1)^3$, so $x=y=z=1$ is the only solution.
Note that Viète's formulas are valid even in the complex plane; the specification of "complex" in the question precludes using real-line tricks like the $x^2\ge0$ inequality in your attempt.