Let $f: X \to Y$ be bijective, and let $f^{-1}: Y \to X$ be it's inverse. Conclude that $f^{-1}$ is also invertible.
Suppose that $f^{-1}(f(x)) = f^{-1}(f(x')) \nRightarrow x=x'$ (not injective), then $x=x' \nRightarrow x=x'$ which is a contradiction. Hence it is injective.
For any $x$ there exists an $f(x)$. Suppose that there exists an $x$ such that $\nexists x' \in X: f^{-1}(f(x'))=x.$ But that means that for some $x$, $\nexists x'\in X: x'=x$. But that $x'$ is simply $x$. This means that for every $x$, there is a corresponding $x$ value that satisfies surjectivity.
Injectivety:
Your solution is not correct, you have to suppose $f^{-1}(x)=f^{-1}(x')$ and not $f^{-1}(f(x))=f^{-1}(f(x'))$. Here is correct procedure:
Suppose we have $x$ and $x'$ such that $f^{-1}(x)=f^{-1}(x')$. Then we have: $$f(f^{-1}(x))=f(f^{-1}(x'))\implies x=x'$$ and thus a conclusion.
Surjectivity:
There is no need to suppose not existence of $x'$ and involving $f$ in first place with $f^{-1}(f(x')) =x$. Remember you have $b$ and you have to find $a$ such that $f^{-1}(a)=b$. Here is faster solution:
Take any $b$, and let $a=f(b)$. Then $$f^{-1}(a) = f^{-1}(f(b)) =b$$ and thus a conclusion.