From what I understand a derivative may not exist at a given point if the function is not continuous or the right and left side derivatives are not equal.
Does that imply that if a function is continuous, the one sided derivatives exist at it's every point?
[Math] Does continuity imply existence of one sided derivatives
calculusderivativesreal-analysis
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There are various ways to explain this, but probably the best way to start is to try to think of "continuous at a point" or "limit at a point" as being its own independent concept, rather than something defined in terms of right-continuity and left-continuity. In the case of the real numbers, there are a lot of ways to define this, but here are two good ones:
Let $D\subseteq \Bbb R$ and let $f\colon D\to \Bbb R$. Then $f$ is continuous at $c\in D$ iff:
For every $\epsilon > 0$ there is a $\delta >0$ such that for all $x\in D$ such that $|x-c|<\delta$, $|f(x)-f(c)| < \epsilon$.
Whenever $(x_i)$ is a sequence in $D$ that converges to $c$, $(f(x_i))$ converges to $f(c)$.
Edit:
Thus the notion of continuity at an endpoint is perfectly sensible. Also, it's quite reasonable to consider a derivative at an endpoint. I think you may be getting a little confused because there are so many theorems out there that apply when a function is continuous on an interval and differentiable in its interior—it's not that the function can't be differentiable there, but that the theorem doesn't need it to be.
As indicated by @levap in the comments, you are likely confusing the (right-hand side, RHS) derivative at $0$ with the (RHS) limit of the derivative. I will just add details to clarify. Assuming you mean $f(0)=0$.
The derivative at $0$ is by definition
$\displaystyle\lim\limits_{h\to0}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0}\frac{h^2\sin(\frac1h)-0}h =
\lim\limits_{h\to0}h\sin(\frac1h)=0$.
Similarly the RHS derivative at $0$ is $\displaystyle\lim\limits_{h\to0^+}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0^+}\frac{h^2\sin(\frac1h)-0}h = \lim\limits_{h\to0^+}h\sin(\frac1h)=0$,
and the LHS derivative at $0$ is
$\displaystyle\lim\limits_{h\to0^-}\frac{f(0+h)-f(0)}h = \lim\limits_{h\to0^-}\frac{h^2\sin(\frac1h)-0}h =
\lim\limits_{h\to0^-}h\sin(\frac1h)=0$.
On the other hand when $x\not=0$ we could use differentiation formulas (apart from the limit definition of the derivative, which we could also use but it would be unnecessary), to obtain that
$f'(x)=2x\sin(\frac1x)-\cos(\frac1x)$, $x\not=0$.
Now, it is true that the limits $\lim\limits_{x\to0}f'(x)$, as well as $\lim\limits_{x\to0^+}f'(x)$ and $\lim\limits_{x\to0^-}f'(x)$ do not exist, that is the derivative $f'(x)$ has no limit (no RHS limit, no LHS limit) as $x\to0$, but this is a different matter than the derivative (the RHS derivative, the LHS derivative) of $f$ at $0$. The latter, as we saw earlier, exists and equals to $0$.
To illustrate, I enclose a couple of graphs (different zoom). $f(x)$ is shown in black, squeezed in between grey $x^2$ an $-x^2$. It should be clear from the picture that the tangent line at $0$ exists and has slope $0$. On the other hand in every neighborhoos of $0$ the slope of $f$ varies between $\approx\pm1$. That is, the dirivative $f'(x)$, shown in red, varies between $\approx\pm1$ (when $x$ is close to $0$ but different from $0$), and hence the limit of the derivative does not exist, as $x\to0$.
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Best Answer
For an elementary example, consider
$$f(x) = \begin{cases} \displaystyle x\sin\frac1x & \text{if } x \neq 0, \\ 0 & \text{if } x = 0.\end{cases}$$
This is obviously continuous, but can't be differentiable on either side of $0$, because the function has points on lines $y=x$ and $y=-x$ arbitrarily close to $0$.