$:y-{{y}_{{}^\circ }}=\frac{-1}{{y}'(3)}(x-{{x}_{{}^\circ }})$  ${{x}_{{}^\circ }}=3\Rightarrow f(3)=0\Rightarrow A=(3,0)$  ${f}'(x)=({{x}^{2}}-2x-3{)}'\sqrt[3]{{{(x+5)}^{2}}}+({{x}^{2}}-2x-3)(\sqrt[3]{{{(x+5)}^{2}}{)}'}$  ${f}'(3)=(2\times 3-2)\sqrt[3]{{{8}^{2}}}+0=4\times 4=16$  $y-0=-\frac{1}{16}(x-3)\Rightarrow 16y=-x+3\Rightarrow 16y+x=3$