I managed to manipulate the expression by changing
$$\begin{align*} |z-3+4i|^4 &= |x + iy - 3 + 4i|^{4}\\ & = (x-3+i(y+4))^2(x-3-i(y+4))^2\\ & = (z-3+4i)^2( \overline{z}-3-4i)^2 \end{align*}$$
I'm not sure where I should go from here. I'm aware I can find two curves on which the function value differs while approaching the limit to prove it doesn't exist, but I can't seem to think of the two curves.
Note that\begin{align}\frac{\left(\overline z-3-4i\right)^4}{\lvert z-3+4i\rvert^4}&=\frac{\ \overline{\left(z-3+4i\right)}^{\,4}\ }{\left(z-3+4i\right)^2\overline{\left(z-3+4i\right)}^{\,2}}\\&=\frac{\ \overline{(z-3+4i)}^{\,2}\ }{(z-3+4i)^2}.\end{align}Now, see what happens when $z=-3+4i+t$ with $t\in\mathbb R$ and what happens when $z=-3+4i+(1+i)t$ with $t\in\mathbb R$.
We have that
$$w=z-3+4i\implies \bar w=\overline{z-3+4i}=\overline{z}-3-4i$$
therefore
$$\lim_{z \rightarrow 3-4i}\frac{(\overline{z}-3-4i)^4}{|z-3+4i|^4}=\lim_{w \rightarrow 0}\frac{\bar w^4}{|w|^4}=\lim_{w \rightarrow 0}\frac{\bar w^4}{w^2\bar w^2}=\lim_{w \rightarrow 0}\frac{\bar w^2}{ w^2}$$
now consider
$w=t \to 0$
$w=(1+i)t \to 0$
