prove that $f(A) \backslash f(B) \subseteq f(A\backslash B)$ 
Let $A,B$ be two subsets of a set $X$,and let $f:X \to Y$ be a function.Prove that $f(A) \backslash f(B) \subseteq f(A\backslash B)$

So I first show what these sets are.
The set $f(A) \backslash f(B) =\{y \in Y \mathtt{\ and \ } \exists x \in B (\ y=f(x)) \mathtt{\ and \ } \neg(\exists u \in B (\ y=f(u)))\}=\{y \in Y \mathtt{\ and \ } \exists x \in B (\ y=f(x)) \mathtt{\ and \ } (\forall u \in B (\ y\ne f(u)))\}$
And the set $f(A \backslash B)=\{y \in Y \mathtt{\ and \ } \exists x \in A\backslash B (y=f(x)) \} =  \{y \in Y \mathtt{\ and \ } \exists x (x \in A \mathtt{\ and \ } x \notin B  \  y=f(x)) \}  $
and  to prove every $y \in f(A)\backslash f(B) \implies y \in  f(A\backslash B) $.
Suppose $y \in f(A)\backslash f(B) $,then exists an $x \in A$ such that $y=f(x)$,and $x$must $ \notin B$ for if $x \in B$ then $f(x)\neq y $ leads to contradition.This means  $y=f(x)$ and $x \in A $ and $x \notin B $ then $y$ is an element of $f(A\backslash B)$.And this complete the proof.
I'm a self-study guy and I feel my statement quite strange.Although I got some intuition about the question,I'm not sure whether I'm doing right. Any help to correct or improve my statement is welcome.
Thanks.
 A: Your proof is almost right, but I don't understand what you mean when you write "for if $x\in B$ then $f(x)\neq y$." This part of the proof should say: for if $x\in B$, then $y=f(x)\in f(B)$, and so $y\notin f(A)\backslash f(B)$, which is a contradiction.
Also, a piece of advice: to develop intuition in problems that show how functions interact with set operations (such as unions, intersections, complements, etc.), it's often useful to think about the reverse inclusion.
So is it also true that $f(A\backslash B)\subseteq f(A)\backslash f(B)$?
The answer is no. Set $A=\left\{0,1\right\}$, $B=\left\{0\right\}$. Then $A\backslash B=\left\{1\right\}$. 
If we let $f(x)=1$ for $x=0$ and $x=1$, then $f(A\backslash B)=\left\{1\right\}$, but $f(A)=f(B)=\left\{1\right\}$, so $f(A)\backslash f(B)=\emptyset$. In other words, even if $x$ isn't in $B$, something in $B$ could map to the same thing that $x$ gets mapped to.
A: Well, to prove that $y \in f(A)\backslash f(B) \Rightarrow y \in f(A\backslash B) $ you take that $y$, then say 
\begin{equation*}\begin{split}
& y \in f(A) \Rightarrow \exists \; a \in A : f(a)=y \\
& y \not\in f(B) \Rightarrow \not\exists \; b \in B : f(b)=y \\
&i.e. \forall b \in B, f(b) \neq y 
\end{split}\end{equation*}
Then $a \in A\backslash B$, therefore $y=f(a) \in f(A\backslash B)$
Basically you did well, I just tried to put it in a more clear form to clarify your not specified doubts
