Proof

Since $f$ is a group homomorphism, it follows for the identity $e_G$ of the group $(G,\ast)$ that $f(e_G)=f(e_G\ast e_G)=f(e_G)\cdot f(e_G).$
The multiplication (inside the group $H$) of the equation with the element and $f(e_G)^{-1}$ gives us $f(e_G)\cdot f(e_G)^{-1}=f(e_G)\cdot f(e_G)\cdot f(e_G)^{-1}.$
This can be simplified to $e_H=f(e_G)\cdot e_H=f(e_G).$

In order to show $f(x^{-1})=f(x)^{-1}$, we multiply the equation (inside the group $H$) with $f(x)$, use the homomorphism property of $f$ and the result in (1), which gives us $f(x)\cdot f(x^{-1})=f(x\ast x^{-1})=f(e_G)=e_H.$
After multiplying both sides of the equation with $f(x)^{-1}$, we get $f(x)^{-1}\cdot f(x)\cdot f(x^{-1})=f(x)^{-1}\cdot e_H.$
This can be simplified to $e_H\cdot f(x^{-1})=f(x)^{-1},$ and finally to $f(x^{-1})=f(x)^{-1}.$
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