Which one is reflexive, symmetric and transitive

a) It's not reflexive but symmetric because addition is commutative: $x+y=0\iff y+x=0$. It's not transitive: for that it's enough to show up a specific counterexample, like $(2,-2)\in R,\ (-2,2)\in R$ while $(2,2)\notin R$ where $R$ denotes the relation defined in the exercise.

b) I think it's meant as $R=\{(x,y):x=y$ or $x=-y\}$. It's reflexive, symmetric and also transitive. To see these more directly we can restate $R=\{(x,y):|x|=|y|\}$.

c) It's not reflexive, not symmetric ($(1,2)\in R$ but $(2,1)\notin R$), and not even transitive: $(1,2),\ (2,4)\in R$ but $(1,4)\notin R$.

d) It's reflexive: for every real number $x$ we have $x\cdot x\ge 0$. It's also symmetric because multiplication is commutative, but it's not transitive: can you find three elements to witness this like above?

e) It's not reflexive: e.g. $(1,1)\notin R=\{(x,y):xy=0\}$ but symmetric, and also not transitive.

f) It's very similar to e) because $xy=0\iff x=0$ or $y=0$. Now $(1,1)$ happens to be in $R=\{(x,y):x=1$ or $y=1\}$ but no other $(x,x)$ pair, e.g. $(2,2)\notin R$.

I'll just give some feedback on a) ... that should help you with the others as well:

a) x + y = 0
This one is reflexive only for (0,0) but not for all real numbers, it is symmetric for (0,0) but not for all real numbers, but I dont know how to see if it is symmetrical.

There is no such thing as 'being reflexive for (0,0) but not for others'. If it holds for all (x,x), then it is reflexive, otherwise it is not reflexive. So this one is not reflexive

In general, to figure out whether some relation is symmetrical, ask yourself this: Is it true that whenever you have $(x,y)$, you also have $(y,x)$? So applied to a), that becomes: Is it true that whenever you have that $x+y=0$, you also have $y+x=0$?

For transitivity, the question is: is it true that whenever you have $(x,y)$ and $(y,z)$, that you then also have $(x,z)$? Applied to a): if $x+y=0$ and $y+z=0$, does that always mean that $x+z=0$?

In (a), the relation $S=\{(x,y)\in\mathbb R^2 : x+y=0\}$, is not reflexive because if $x \neq 0$ then $x+x \neq 0$.
It is symmetric because if $x+y=0$, then $y+x=0$.
It isn't transitive, because (for example) $(1,-1) \in S$, $(-1,1) \in S$, but $(1,1) \notin S$.

In (b), the relation is $T=\{(x,y) \in \mathbb R^2 : x = \pm y\}$.
It is reflexive because $x=x$ for all $x \in \mathbb R$.
If $x = \pm y$ then $y = \pm x$, whence it is symmetric.
Also, if $x = \pm y$ and $y = \pm z$ then $x = \pm z$, so it is transitive.

Now try to prove that in (c), $R = \{(x,y) \in \mathbb R^2: x=2y\}$ is not reflexive (aren't there numbers which are not the double of themselves?), not symmetric (aren't there numbers $x$ and $y$ such that $x=2y$ but $y \neq 2x$), and not transitive (if $x=2y$ and $y=2z$ then $x=4z$; there are numbers $z$ for which $2z \neq 4z$, right?).

The relation in (d) is reflexive because $x^2 \geq 0$ for every $x$; symmetric because $xy=yx$, so if $xy\geq0$... It's not transitive: $(1,0)$ and $(0,-1)$ are in the relation, but not $(1,-1)$.

In (e) it's not reflexive, symmetric and not transitive.

In (f) the relation is clearly not reflexive (think of $(0,0)$), it's symmetric but not transitive.

I think you should take a look at the definition of these 3 possible properties of relations on $\mathbb{R}$ again. The relation $x\sim y \iff x+y = 0 $ is indeed symmetric, since $x+y = 0 \iff y+x = 0$. I don't think this is the place to spoil all the answers, taking away from you the chance to think about it again.