Is there a function $f$ with $f'(x) = e^x * f(x)$? Edit : I'm ''derivation'' and ''integration'' beginner. So I don't have any techniques to solve equations like that.This why I think there have to be a clever trick to solve this. a), b), c) was okay. You can see my solution below. I want to emphasize that I don't want to see the solution. Maybe someone can show me the techniques for $b)$? I found a $f$ in b), c) only with try and error. If you show me the trick, I would  try again $c)$ and $d)$ by my own.

Are there any functions $f$ with following equations:

$ a) f'(x) =e^x + f(x)   $  a) is clear now. Thank you for all of your help

$ b) f'(x) =e^x \cdot f(x) $ solved BUT only with try and error. what is the beginner technique ( like I said: Im a beginner ) to find $f$?

$ c) f'(x) =e^x \cdot f(x)^2 $ solved BUT only with try and error. what is the beginner technique to find $f$?

$ d) f'(x) =e^{f(x)} $

Remark : $f$ is not a constant function.

Update: $a ) f:= x \cdot e^x $. Then we use the multiplication rule for derivation. $f'$ = $x \cdot e^x + 1 \cdot e^x = x \cdot e^x + e^x$. 

Update2 b) $f:= c \cdot e^{e^x}.$ Then we can use the chain rule: $f' = c \cdot e^{e^x} \cdot e^x$.

Update3 c) $f:= - \frac{1}{e^x +c}$. Then use chain rule and quotient rule: $ f' = e^x \cdot  \frac{1}{(e^x +c)^2}$
$$f′(x)=e^x+f(x)\Longleftrightarrow f'-f =e^x\Longleftrightarrow(fe^{-x})'=1\Longleftrightarrow fe^{-x}=x-c\Longleftrightarrow f(x)= e^x(x+c)$$
In order to be sure whether such a function exist you would need to solve the differential equations. b),c) and d) can be solved by separation of variables. a) is on the form $$y^\prime(x) + P(x)y(x) = Q(x), \ \ P(x) = -1,  \ \ Q(x) = e^x$$
and can thus be solved by the method of integrating factor https://en.wikipedia.org/wiki/Integrating_factor.