Помогите пожалуйста решить алгебру

Решите задачу:

$1)\quad f(x)= \frac{x^2-3x}{x-4}\\\\f'(x)= \frac{(2x-3)(x-4)-(x^2-3x)\cdot 1}{(x-4)^2} = \frac{2x^2-7x+12-x^2+3x}{(x-4)^2} = \frac{x^2-4x+12}{(x-4)^2} \\\\f'(6)= \frac{6^2-24+12}{2^2} = \frac{0}{4}=0\\\\2)\quad f(x)=x-\frac{1}{x}\\\\f'(x)=1+\frac{1}{x^2}\\\\x\cdot f'(x)=f(x+4)\; \; \Rightarrow \; \; x\cdot (1+\frac{1}{x^2})= (x+4)-\frac{1}{x+4}\\\\ \frac{x(x^2+1)}{x^2} = \frac{(x+4)^2-1}{x+4} \; ;\; \; x\ne 0\; ,\; x\ne -4$

$(x^2+1)(x+4)=x(x^2+8x+15)\\\\x^3+4x^2+x+4=x^3+8x^2+15x\\\\4x^2+14x-4=0\\\\2x^2+7x-2=0\\\\D=49+16=65\\\\x_{1}= \frac{-7- \sqrt{65}}{4} \; ;\; x_2= \frac{-7+\sqrt{65}}{4}$