I know this exercise can be solved in several ways, such as Ravi Vakil's proof or this answer, but I'd like to try to follow the given hint if possible.
The exercise says the following:
Let $f:X→Y$ be a dominant morphism of integral schemes of finite type over a field $k$. Show there is an open dense subset $U⊆X$ s.t. $\dim U_y=\dim X−\dim Y$ for all $y$ in the image of $U$.
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=SpecA$ and $Y=SpecB$. Then $A$ is a finitely generated $B$-algebra. Take $t_1,\dots,t_e∈A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=SpecB[t_1,...,t_e]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X→X_1$ is generically finite. Now use (Ex. 3.7) above.
I've done most of the process, so I'll go directly to the place I'm stuck. Let $g:X\to X_1$ the morphism of the hint, which I've proved that is dominant, generically finite and of finite type.
By Ex II.3.7 there exists an open dense subset $U_1\subseteq X_1$ such that, takin $U=g^{-1}(U_1)$, the induced morphism $g|_U:U→U_1$ is finite. I claim that $U⊆X$ is the desired set. Certainly it is a dense open set, since $X$ integral implies $X$ is irreducible (Proposition 3.1) and $U$ is nonempty, so that it is automatically dense. Thus, we need only check that $y∈f(U)$ implies $\dim U_y=e$.
I think using (b) I can show that $\dim U_y\geq e$, by I don't know how to proof the other inequality. In addition I haven't (at least explicitily) used that $X_1$ is an affine $e$-space, so that might be the key.
Part (b), under the same hypothesis as (c)
Let $e=\dim X-\dim Y$. For any point $y\in f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $\geq e$.