Proving Vector Subspaces Question 1:
The set $\mathbb R^3$ of all column vectors of length three, with real entries, is a vector space. Is the subset $$B=\{xyz \in \mathbb R^3 \mid xy+yz= 0\}$$ a subspace of $\mathbb R^3$? 
Justify your answer.
Attempted answer:
$(0)$: let $0$ vector be in set $B$, then $0\cdot0 + 0\cdot0 = 0$, hence $B$ is non-empty.
$(1)$ A1: Let $u = \begin{bmatrix}0&0&1\end{bmatrix}$ and $V = \begin{bmatrix}0&1&0\end{bmatrix}$, which are both in $B$, but $u +v = \begin{bmatrix}0&1&1\end{bmatrix}$ is not in $B$, since $0\cdot1 +1\cdot1 \neq 0$. Hence not closed under multiplication.
$(2)$ S1: I know its irrelevant since its not closed under A1...but is this correct--> let $u = \begin{bmatrix}a&b&c\end{bmatrix}$, then $k\cdot u = \begin{bmatrix}ka&kb&kc\end{bmatrix}\implies (ka)(kb) + (kb)(kc) = 0 \implies k^2(ab) + k^2(bc) = 0$, hence must clearly be in set $B$?
Question 2
Show that the set of all twice differentiable functions $f:\mathbb R\to \mathbb R$ satisfying the differential equation $\sin(x)f''(x) + x^2f(x) = 0$ is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Note that $f''$ is the second derivative of $f$.
Attempted answer: The zero vector in $F$ space has the zero function $0(x) = 0 \in \mathbb R$ and we have $0''(x) = 0'(x) = 0$. Assume that the set of solutions for the equation is a subset of $F$ space, hence it contains the zero function and so $0(x) = \sin(x)\cdot 0''(x) + x^2\cdot0(x) = 0+0 = 0$ so set of solutions is in subspace.

Question 1) 
To prove U (some arbitrary subspace) is a subspace of V (some arbitrary vector space) you need to prove  a) the zero vector is in U b) U is closed by addition c) U is closed by scalar multiplication by the field V is defined by (in your case any real number) d) for every $ u \in U$, $u \in V$.  
a) Obviously true since when x, y, z = 0, xy + yz = 0. And {0,0,0} is the zero vector in $R^3$.
b) Show that if $v_1 = {x_1, x_2, x_3}$ and $v_2= {y_1, y_2, y_3}$ then $x_1+y_1, x_2+y_2, x_3+y_3$ satisfies xy + yz = 0.
c) Show if $v_1= {x_1, x_2, x_3}$ then $\lambda$$x_1$, $\lambda*x_2$, $\lambda*x_3 \in U$ (or in other words satisfies xy+yz= 0).
d) This is obviously true.
Question 2) Do the same for number 2.  The idea that the derivative function is a linear function is abundantly shown with quick googling.
