A mode represents the same quantity in continuous distributions and discrete distributions: The element in a random variable's domain at which the pdf is maximized. Sure, for continuous distributions you have to fudge the end of that a bit to something like "at which the pdf is locally maximized," but it's the same principle. In other words, your reasoning is valid.
In this case, our domain is the closed interval $[0,1]$, so the pdf $ 3x^{2}$ takes on a maximal value at either a critical point or at the endpoints $0,1$. The only critical point is $0,$ and $3x^{2}_{x=0} = 0$. Since $3x^{2}>0$ at $x=1$, your answer is correct: The mode is $1$.
Regarding how to interpret the meaning behind a mode of real-valued random variable, I wouldn't analyze it too much yet. ;) The bizarre, seemingly paradoxical idea of a real-valued random variable having zero probability at any isolated point can be resolved.
But it takes some analysis and topology to really get comfortable with that idea. After some of those classes, you can gleefully look back and interpret the idea more clearly and intuitively.
Best Answer
For strictly continuous random variables the value $m$ is unique and always in the support (the set of possible values) of $X$, thus there is no problem to define it as $P(X\ge m) = 0.5$. In discrete random variable whether with finite or infinite countable support, it is possible that no possible value of $X$ will satisfy $P(X\ge m )=0.5$. Let us consider a simple example. Let $X \sim \mathcal{B}in (3, 1/2)$, hence we have to find $m$ such that $ P(X \le m ) =0.5 $. Let us check possible value of $m$ $$ P(X\le 1) = \frac{1}{2^3} + \frac{3}{2^3} = 1/2. $$ Bingo! $1$ is the median. Now, take $p=1/4$ instead of $1/2$, $$ P(X\le 1) = \frac{3^3}{4^3} + \frac{3}{2^3} = 0.74, $$ that is way too high probability, however $$ P(X\le 0) = \frac{3^3}{4^3} = 0.42. $$ That is too low. I.e., there is no value that $X$ can get and it will satisfy $ P(X\le m) = 0.5, $ hence you have to chose $0$ or $1$ as the median. Any value in $(0,1)$ will not change a thing as $X$ cannot receive these values.