Ah, found a proof myself :), is this correct?
Take $x \in U \cap W$. Since $x \in U$ it holds that $x = \lambda (2-e^{3x})+ \mu (4x + e^3x)$. $x \in W \Rightarrow x = \alpha \cdot 1 + \beta e^x + \gamma e^{3x}$. For any $\gamma, \mu, \alpha, \beta, \gamma$. Since $x = x$ we can subtract those two expressions which gives: $0=1 \cdot (2 \lambda - \alpha) + x(4 \mu) - e^x \cdot \beta + e^{3x} (\mu -\lambda - \gamma)$ and since $1, x, e^x$ and $e^{3x}$ are linear independent (I will not proof that), it follows that $\beta = 0, \mu = 0, 2 \lambda = \alpha, -\lambda=\gamma$. Now choose $\epsilon = \lambda$, then $\gamma = -\epsilon$ and $\alpha = 2 \epsilon$. Filling this in for $x$ gives $x = \epsilon (2 - e^{3x})$ and in the other equation: $x = 2\epsilon - \epsilon e^{3x} = \epsilon(2-e^{3x})$. And since $\lambda$ is a free variable, so is $\epsilon$, so $x \in U \cap W \Leftrightarrow x \in <2 - e^{3x}>$.