If $y = f(x)$ is symmetric about the lines $3x+4y+1=0$ and $4x-3y-7=0$ then prove that it must be symmetric about $(1,-1)$.
Can we comment on symmetricity of a function about point of intersection of two non perpendicular lines (given that the function is symmetric about those two line)?
Let $\Gamma$ denote the graph of $f$: $\Gamma =\{(x,y)\in\Bbb R^2 :y=f(x)\}$ (more generally, $\Gamma$ could be any set of points in the plane $\Bbb R^2$).
If $A, B$ are transformations of the plane (or arbitrary functions $\Bbb R^2 \to \Bbb R^2$), which both leave $\Gamma$ fixed, i.e. $A(\Gamma) :=\{A(p):p\in\Gamma\}\ =\Gamma=B(\Gamma)$, then so does the composite transformation $B\circ A$: $$B\circ A\, (\Gamma) =\{B\circ A(p):p\in\Gamma\} =\{B(A(p)) :p\in\Gamma\}=B\big(\{A(p):p\in\Gamma\} \big) =B\big(A(\Gamma)\big)=B(\Gamma) =\Gamma$$
In this example, you can explicitly write up the coordinate mappings belonging to the two reflections, and then verify that their composition (in any order) is the reflection about the intersection point $p_1=(1, -1)$: $$R_{p_1}=(x,y)\mapsto (2-x,\, -2-y)$$