First, simplify your $f'(x)$ before taking the second derivative.
$$f'(x)= \dfrac{(\sqrt{25-x^2})^2 - (x^2)}{\sqrt{25-x^2}} $$
$$ = \dfrac{25 - 2x^2}{\sqrt{25-x^2}}$$
Now use the quotient rule to compute $f''(x)$.
(You need to correct your $f''(x)$ and simplify, to find the common denominator), then determine when $f''(x) = 0$. That's where an inflection point would be.
Of course, we also need to check $x = 5, x = -5$ because that is where the denominator will be zero, and the derivatives (first and second) undefined. So just check what is happening there $f(0)$, however, will be zero as well.
$$f''(x) = \frac{(-4x)\sqrt{25-x^2}-(25-2x^2)\large\frac{-x}{\sqrt{25-x^2}}}{25-x^2} $$
$$ = \frac{(-4x)(25-x^2)-(25-2x^2)(-x)}{(25-x^2)\sqrt{25-x^2}} $$
$$ = \frac{-75x+2x^3}{(25-x^2)^{\large\frac32}} $$
$$ = \frac{x(2x^2 - 75)}{(25-x^2)^{\large\frac32}} $$
$$ = \frac{2x(x^2 - \frac{75}{2})}{(25 - x^2)^{\large \frac 32}}$$
Now, when is $f''(x) = 0$? Be careful with $x = 5, x= -5$: you want to check the point, but note that neither the first nor the second derivative is defined there.
When $2x = 0,\; f''(x) = 0$, as is $f'$ and $f$.
When $x^2 - 75/2 = 0, f''(x) = 0 \implies x \pm 5\sqrt{\frac 32} = 0 \implies $
But to confirm that that there is an inflection point at $x=0$, you'll want to check whether $f''(x)$ changes signs on there, from negative to positive, or positive to negative. It does, indeed change signs at $x = 0$: as $x \to 0$ from the left, $f''(x) > 0$, and when $x \to 0$ from the right, $f''(x) <0$. So there is, indeed, an inflection point at $(0, 0)$.
Take for example $$
f(t) = \begin{cases}
-x^2 &\text{if $x < 0$} \\
x^2 &\text{if $x \geq 0$.}
\end{cases}
$$
For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.
For your second question, maybe things are clearer if stated like this
If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point
This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.
But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.
Best Answer
As you stated in the comments, the second derivative is $$y'' = 6ax + 6 + e^x = 0$$
First, we note that if $a=0$ the equation becomes $$ 6 + e^x = 0$$ which has no real solutions, thus there are no inflection points.
Rewriting the equation as $$x = -\dfrac{6+e^x}{6a}$$ or $$x = -\dfrac1{a} \cdot \left(1 + \dfrac 16 e^x \right)$$ makes the behavior clearer.
If $a \gt 0$ we can see that $x$ must be negative and thus $e^x < 1$ and there will be a solution in the neighborhood of $x = -\dfrac1a$.
I will leave the case where $a \lt 0$, thus $x \gt 0$ for you to consider.