How can I show that Xl$X_l$ and Xf∘l$X_{f\circ l}$are homotopy equivalent?

This is not true. Let X=R2$X={\mathbb{R}}^2$ and f:R2→R2$f:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be defined as f(x)=0$f(x)=0$ for all x∈R2$x\in {\mathbb{R}}^2$. Define H:R2×[0,1]→R2$H:{\mathbb{R}}^2\times [0,1]\to {\mathbb{R}}^2$ as H(x,t)=tx$H(x,t)=tx$ for all (x,t)∈R2×[0,1]$(x,t)\in {\mathbb{R}}^2\times [0,1]$. Then, H:f≃idR2$H:f\simeq {\text{id}}_{{\mathbb{R}}^2}$.

Next, let l:[0,1]∋t⟼e2πit∈S1$l:[0,1]\ni t⟼e^{2\pi it}\in {\mathbb{S}}^1$ be the loop based at 1∈S1$1\in {\mathbb{S}}^1$. Note that Xl=S1$X_l={\mathbb{S}}^1$ and Xf∘l={0}$X_{f\circ l}=\{0\}$. But, S1${\mathbb{S}}^1$ is not homotopically equivalent to {0}$\{0\}$ as π1(S1)=Z${\pi }_1({\mathbb{S}}^1)=\mathbb{Z}$ but π1({0})=trivial group${\pi }_1(\{0\})=\text{trivial group}$.