I think you are talking about
$$
\langle x,y \rangle=\langle x,z \rangle
$$
being true for any $x,y,z \in (V, \langle\cdot,\cdot\rangle)$, then you can not infer $y$ and $z$ being parallel to each other.
A counterexample would be $V=\mathbb{R}^3$ and $x=(1,0,0)^t, y=(0,1,0)^t$ and $z=(0,0,1)^t$. Even though with the standard inner product
$$
\langle x,y \rangle=1\cdot0+0\cdot1+0\cdot0=0=1\cdot0+0\cdot0+0\cdot1=\langle x,z \rangle
$$
your identity holds, it is obviously not true that $y$ is parallel to $z$.
However, the following holds because of the bilinearity of the inner product:
$$
\langle x,y \rangle=\langle x,z \rangle \Leftrightarrow \langle x,y-z \rangle=0
$$
which means that $x \perp y-z$.
I think you are talking about $$ \langle x,y \rangle=\langle x,z \rangle $$ being true for any $x,y,z \in (V, \langle\cdot,\cdot\rangle)$, then you can not infer $y$ and $z$ being parallel to each other. A counterexample would be $V=\mathbb{R}^3$ and $x=(1,0,0)^t, y=(0,1,0)^t$ and $z=(0,0,1)^t$. Even though with the standard inner product $$ \langle x,y \rangle=1\cdot0+0\cdot1+0\cdot0=0=1\cdot0+0\cdot0+0\cdot1=\langle x,z \rangle $$ your identity holds, it is obviously not true that $y$ is parallel to $z$. However, the following holds because of the bilinearity of the inner product: $$ \langle x,y \rangle=\langle x,z \rangle \Leftrightarrow \langle x,y-z \rangle=0 $$ which means that $x \perp y-z$.