Injective and surjective function of $~f(x,y)~$. 
Consider a funstion $~f:\mathbb R^2\to \mathbb R~ $ such that $~f(x,y)= x^2+y^3~$. Evaluate the injective and surjective function of $~f(x,y)~$.

I substitute $~f(x,y)~$ with a value of $~0~$, then observe the graph of $~x^2+y^3=0~$.
There are two values of $~x~$ that correspond to $~y=-0.5~$. Therefore the function is not injective (I believe this is correct)
However, for the surjective part, which is defined as for every value of $~y~$, there exist a corresponding $~x~$ value (need not to be one on one). I am confused here. If a function is continuous, that function is surjective, as there isn't any break along the graph that is 'broken'.
My interpretation for $~f(x,y)~$ is that it is not surjective, because of the $~x^2~$ term? 
And finally, is $~f(x,y)~$ surjective and injective?
 A: If you have a continous function $f : \Bbb{R} \to \Bbb{R}$ (note the domain is not $\Bbb{R}^2$), then you get the intermediate value theorem. Specifically, if $a$ and $b$ are in the range of $f$, then so is every point between $a$ and $b$. So, if the function tends to $\infty$ on one end, and $-\infty$ on the other, it's not hard to argue that the function is surjective (which is what I think you're thinking of).
Here, it's a little more complicated, since the domain is $\Bbb{R}^2$. However, we can indeed simplify it to a function of one variable, by simply choosing (arbitrarily) the value of one of the variables. Remember, the range is the set of all possible values of $f(x, y)$, so we have complete control over the values of $x$ and $y$ when we try to produce a given value.
In particular, consider what happens when $x = 0$. We get $f(0, y) = y^3$. This is now a function from $\Bbb{R}$ to $\Bbb{R}$ that is indeed surjective. You can either argue it the same way as above, or simply observe that, for any $r \in \Bbb{R}$, we have
$$f(0, \sqrt[3]{r}) = r.$$
That is, for any point $r$ in the codomain $\Bbb{R}$, we have a point $(0, \sqrt[3]{r})$ in the domain $\Bbb{R}^2$ that maps to $r$. This is the very definition of surjectivity.
