= Solution
A two-dimensional <Lie algebra representation> already provides a counterexample. In the basis $e_1,e_2$ of $V=\mathbb C^2$, define
$$
\rho(x)=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad
\rho(y)=\rho(z)=0.
$$
The commutator $[\rho(x),\rho(y)]$ is zero, agreeing with $\rho(z)$; the other defining <Lie brackets> also map to zero. Thus $\rho$ is a <Lie algebra homomorphism>, although it is not a <Faithful Lie algebra representation>.
The line $L=\mathbb Ce_1$ is an <invariant subspace>. Any complementary line has the form $\mathbb C(e_2+a e_1)$ for some $a\in\mathbb C$. But
$$
\rho(x)(e_2+a e_1)=e_1\notin\mathbb C(e_2+a e_1),
$$
so no complementary line is an <invariant subspace>. Equivalently, every invariant line is killed by $\rho(x)$: its restriction to a line is a scalar, and its square is zero, forcing that scalar to be zero. A decomposition into two invariant lines would then force $\rho(x)=0$, a contradiction. This representation is therefore not a <semisimple representation>. Thus
$$
\boxed{\text{Not every finite-dimensional representation of }\mathfrak h\text{ is completely reducible}.}
$$
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