I'm going to use $f$ and $g$ for the functionals. Something about using $x$ and $y$ to stand for completely different objects bothers me.
I'm assuming that $f(x) = 0$ whenever $g(x) = 0$, so if we plug $x$ into $g$ and get $0$, we must also get $0$ when plugging the same $x$ into $f$ - but there may be $x$s we can plug into $f$ to get $0$ which don't evaluate to $0$ when plugged into $g$. The goal is to show that $f(x) = \alpha g(x)$ for some scalar $\alpha$.
First a stupid case: If $g(x) = 0$ for all $x$, then $f$ and $g$ are both the $0$ functional, so are multiples of each other with any $\alpha$.
So, we may assume there is an $x$ with the property that $g(x)\neq 0$. The rank-nullity theorem asserts that the kernel $\ker g$ of $g$ is a codimension $1$ hyperplane in the vector space. Choose a basis $\{w_1,w_n,...\}$ for this kernel and extend this basis to the set $\{w_1,..., x\}$. I claim that this new set is a basis for the whole vector space. Since it has the right number of elements, we must only show it's independent.
So, assume $\sum c_i w_i + dx = 0$. Applying $g$ to both sides gives $\sum c_i g(w_i) + dg(x) = 0$. Since each $w_i\in\ker g$, we get $dg(x) = 0$. This implies $d = 0$. Since the $w_i$ are assumed to be independent, this implies all the $c_i$ are $0$, so we really do have a basis.
Now, define $\alpha = f(x)/g(x)$. I claim that $f = \alpha g$.
To see this, notice that given any linear combination of $w_i$ and $x$, $\sum c_i w_i + dx$, we have $g(\sum c_i w_i + dx) = dg(x)$ while $f(\sum c_i w_i + dx) = df(x) = d\alpha g(x)$.
So $f(\sum c_i w_i + dx) = \alpha g(\sum c_i w_i + dx)$ and since $\{w_1,...,w_n, x\}$ forms a basis, we're done.