Let $X=(C^1([0,1],\mathbb{R}), \|\cdot\|)$ with $\|f\|= max\{\|f\|_{\infty}, \|f^\prime\|_{\infty}\}$ be given. Moreover, let $U= \{f \in X |f(0)=0\}$ and consider the map $\varphi \colon U \to \mathbb{R} , f \mapsto \int_{0}^{1} \frac{f(x)}{x} dx$.
I have to show that $\varphi \in U^\prime$ (i.e. $\varphi$ is a linear and continuous map $U \to \mathbb{R}$) with $\|\varphi\|_{U^\prime} =1$. Clearly, $\varphi$ is linear and for $f(x)=x$, we have $|\varphi(f)|=1$ and $\|f\|=1$, so $\|\varphi\|_{U^\prime} \geq 1$ holds. However, I struggle with showing $\|\varphi\|_{U^\prime} \leq 1$.
Furthermore, I have to construct an explicit linear and norm-preserving extension $\Phi \in X^\prime$ of $\varphi$, but I have no idea, how to do that
Using mean value theorem, you have that for all $f$ s.t. $\|f\|=1$, that $$|f(x)|\leq |x|.$$ Therefore $$|\varphi (f)|\leq 1$$ for all $\|f\|=1$, and thus $\|\varphi \|_{U'}=1$.
For $\Phi$, may be $$\Phi(f)=\int_0^1\frac{f(x)-f(0)}{x}\,\mathrm d x \ \ ?$$ You have that $\Phi(f)=\varphi (f)$ for all $f\in U$ and $\|\Phi(f)\|_{X'}=1$.