Vtyjkyuk

How to prove that $$f: Bbb RtoBbb R:xmapsto x^3-3x$$ is surjective? By expressing $x$ in terns of $y$? I am so confused right now.

First, let me say that finding a value $x$ such that $y = x^3 - 3x$ is perfectly possible to do, but really horrible. Cardano's method is how you find roots of a cubic, but the formula that comes out the end is far, far worse than the quadratic formula. I would strongly recommend against using this, unless you really need a precise, specific value of $x$ such that $y = x^3 - 3x$.

Let $f(x) = x^3 - 3x$, and fix some $y in mathbbR$. Since $f$ is a continuous function, it suffices to find $a$ and $b$ such that
$$f(a) < y < f(b)$$
because, by the intermediate value theorem, there must be some $x$ between $a$ and $b$ such that $f(x) = y$ (which is what we need to prove surjectivity).

To find an appropriate $b$, consider first the possibility the $y le 2$. If $y < 2$, then $f(2) = 2$, so we can just pick $b = 2$. (If $y = 2$, then just outright choose $x = 2$, and be done with it!)

Otherwise, suppose $y > 2$. Note that, for $x ge 2$, we have $x^3 ge 4x$, so
$$f(x) = x^3 - 3x ge 4x - 3x = x.$$
Thus, if we just choose $b = y + 1$, then $b > 3 ge 2$, so
$$f(b) ge b = y + 1 > y$$
as required.

We have our upper estimate $b$. To find a lower estimate of $a$, we can use the symmetry of the function. Note that $f$ is odd, that is, $f(-x) = -f(x)$ for all $x$. Therefore, by the reasoning above, if $y = -2$, we can use $x = -2$. If $y > -2$, then set $a = -2$. If $y < -2$, then setting $a = y - 1$ will work.

Either which way, we now have $a$ and $b$ such that $f(a) < f(y) < f(b)$. Because $f$ is continuous, the intermediate value theorem kicks in, and so there exists some $x$ between $a$ and $b$ such that $f(x) = y$.

The way the question is phrased right now is not quite clear, as surjectivity is a property of functions.
Precisely, you mean to say:

How do I show that the function $f:mathbbRrightarrowmathbbR$ given by $f(x)=x^3-3x$ is surjective?

A function $f$ is surjective if for every element $y$ in its codomain, you can find an element $x$ in the domain such that $f(x)=y$.
In your case, both the domain and codomain are $mathbbR$.

Hint: Note that $f(x) rightarrow pm infty$ as $x rightarrow pm infty$ and $f$ is continuous.