Associated with the recurrence $f(n)=-3f(n-1)+4f(n-2)$ is a so-called characteristic equation, $x^2=-3x+4$. Its coefficients are the same as the coefficients of the recurrence, and the powers of $x$ are chosen so that the smallest exponent is $0$, associated with the smallest argument of $f$, which in this case is $n-2$; the exponents then increase in step with the arguments of $f$, so that exponent $1$ goes with $(n-2)+1=n-1$, and exponent $2$ goes with $(n-2)+2=n$.
Now solve the auxiliary equation: $x^2+3x-4=0$, $(x+4)(x-1)=0$, $x=-4$ or $x=1$.
There is a general theorem that says that when the roots are distinct, as they are here, the general solution to the recurrence has the form
$f(n)=Ar_1^n+Br_2^n\;,$ where $r_1$ and $r_2$ are the two roots. Thus, for this recurrence the general solution is $f(n)=A(-4)^n+B\cdot1^n=A(-4)^n+B\;.\tag{1}$
$(1)$ gives all solutions to the recurrence $f(n)=-3f(n-1)+4f(n-2)$, for all possible initial values of $f(1)$ and $f(2)$. To determine which values of $A$ and $B$ correspond to your particular initial values, substitute $n=1$ and $n=2$ into $(1)$. For $n=1$ you get $1=f(1)=A(-4)+B\;,$ and for $n=2$ you get $2=f(2)=A(-4)^2+B\;.$
Now you have a system of two equations in two unknowns,
$\left\{\begin{align*} &-4A+B=1\\ &16A+B=2\;. \end{align*}\right.$
Solve this system for $A$ and $B$, substitute these values into $(1)$, and you have your general solution. (I get $A=\frac1{20}$ and $B=\frac65$.)
Note that if the the roots $r_1$ and $r_2$ of the characteristic equation are equal, say $r_1=r_2=r$, the general solution is a little different:
$f(n)=Ar^n+Bnr^n\;.$ However, you solve for the particular $A$ and $B$ in the same way.