The derivative provides information about the gradient or slope of the curve/graph of a function which can be used to locate points on the function's curve/graph where its gradient is $0$. These points are often associated with the largest or smallest values of the functions.
A stationary point is any point where the tangent to the curve/graph is horizontal. These can be calculated by finding the derivative and equating it to $0$. i.e. stationary points can be located by looking for points at which $\frac{dy}{dx}=0$.
At a turning point, $\frac{dy}{dx}=0$. So all turning points are stationary points. But not all points where $\frac{dy}{dx}=0$ are turning points, i.e. not all stationary points are turning points.
The local maximum can be defined as a point where $f(a)\ge f(x)$ for all $x$ in the interval in layman's terms $f(a)$ is the highest point in the interval. While the local minimum is the opposite, $f(a)\le f(x)$ for all $x$ in the interval aka $f(a)$ is the lowest point in the interval.
Now, the link between all these terms and the derivative can be discovered when you examine the gradient of the curve/graph. Think about the graph of $y=x^2$ (you can just envision a parabola). Clearly at the bottom of the curve $\frac{dy}{dx}=0$, from the left $\frac{dy}{dx}$ is negative (because the slope is going down) and from the right $\frac{dy}{dx}$ is positive (because the slope is going up). $\frac{dy}{dx}$ goes from negative to $0$, to positive and thus is clearly increasing.
The second derivative is used to verify that a point is a local maximum or minimum. If the second derivative is positive then $\frac{dy}{dx}$ would be increasing and that stationary point would be a minimum turning point.
In conclusion, if $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}>0$ at that same point, them the point must be a minimum. This can also be observed for a maximum turning point. If $\frac{dy}{dx}=0$ (is a stationary point) and if $\frac{d^2y}{dx^2}<0$ at that same point, them the point must be a maximum. You can read more here for more in-depth details as I couldn't write everything, but I tried to summarize the important pieces.
A point of inflection on the graph/curve is a point where the gradient stops increasing and starts decreasing, or stops decreasing and starts increasing. What you have is correct, since $\frac{d^2y}{dx^2}=0$, at $(1,\space 1)$ that is indeed an inflection point. As Fly by Night suggested, try youtubing videos, Khan Academy is a good source.
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A colleague of mine from the late 1990s used the term terrace point for a point $x=a$ where the first derivative is zero and the first derivative doesn't change sign as you pass through $x=a,$ and I liked the idea so much that I subsequently used it in all my calculus classes. Of course, this was only used in simple introductory calculus examples where the zeros of the derivative are isolated from each other.
A few years later (in July 2002; see 6. TERRACE POINTS IN THE FIRST DERIVATIVE TEST here) I was told that the term "terrace point" is in Ostebee/Zorn's Calculus book, although I've never looked at a copy to be sure, and I don't know whether the term was in both the 1994 1st and 2002 2nd editions or only in the 2002 2nd edition.
Anyway, when I last wrote about this term (as far as I can recall), there wasn't nearly as much on the internet as there is now, and google-books searching wasn't available. A google search shows that the term "terrace point" is now fairly widely used. Also, a google-books search shows that it is definitely used in the 2002 2nd edition of Ostebee/Zorn's book, as well as in several other books. Interestingly, the term also appears on p. 40 (line −10) of William Richard Ransom's 1915 Early Calculus. As far as I can determine, the only use of "terrace point" in one of the math oriented Stack Exchanges is this answer from 24 October 2019.
One of the reasons I liked having a name for this this notion is that it allows you to label all four of the possibilities that can show up on a first derivative sign chart where the derivative is zero (and isolated from the other zeros):
$$ ++++0++++ \;\;\;\;\; \text {ter} $$
$$ ++++0---- \;\;\;\;\; \text {max} $$
$$ ----0++++ \;\;\;\;\; \text {min} $$
$$ ----0---- \;\;\;\;\; \text {ter} $$