$R=0$ solution in field equations

I am dealing with some General Relativity extensions and I am not sure about my knowledge in basic GR since I am having some weird troubles with what I think are basic concepts.

As far as I know, if we have a field equation (we can think about this as, for example, EFE) in vacuum, $R_{munu}-dfrac{1}{2}g_{munu}R=0$. In a more general sense, think about it as some equation which contains in the left hand side objects like $R$ or $R_{munu}$ and its derivatives, but not higher tensors as Riemann) we can see that the trace of the Einstein equation implies $R=0$, and plugging this into the original equation gives us $R_{munu}=0$. Now, is it correct to say that every solution of $R_{munu}-dfrac{1}{2}g_{munu}R=0$ must have neccesairly $R_{munu}=0$? I think this is true.

Since the Schwarzschild metric has $R=0$, can we conclude that a field equation in the general form I described before verifies Schwarzschild as a solution if Einstein equation $R_{munu}-dfrac{1}{2}g_{munu}R=0$ is verified?

I mean, suppose $D(X,Y)$ denotes some function that can contain derivatives of its arguments, and then we have the field equation $D(R,R_{munu})+R_{munu}-dfrac{1}{2}g_{munu}R=0$. If Einstein is verified, $R_{munu}-dfrac{1}{2}g_{munu}R=0$ and then $D(R,R_{munu})=0$, but since $D(R,R_{munu})$ can only have derivatives of $R$ and $R_{munu}$, and Einstein is verified so $R=0$ and $R_{munu}=0$, of course its derivatives will be zero and $D(R,R_{munu})=0$, so $R=0$ is a solution and since Schwarzschild has $R=0$ it must be a solution to this extended theory. Is my reasoning correct? Thanks