50 баллов помогите пожалуйста

Решите задачу:

$1)\; \; \frac{y}{x}=arctg(xy)\\\\\frac{y'x-x'y}{x^2}=\frac{1}{1+(xy)^2}\cdot (x'y+xy')\; \; \; \; \; [\; x'=1\; ]\\\\\frac{y'x-y}{x^2}=\frac{y+xy'}{1+x^2y^2}\\\\\frac{y'}{x}-\frac{y}{x^2}=\frac{y}{1+x^2y^2}+\frac{xy'}{1+x^2y^2}\\\\y'\cdot \Big (\frac{1}{x}-\frac{x}{1+x^2y^2}\Big )=\frac{y}{1+x^2y^2}+\frac{y}{x^2}\\\\y'=\frac{x^2y+y+x^2y^3}{x^2(1+x^2y^2)}:\frac{1+x^2y^2-x^2}{x(1+x^2y^2)}\\\\y'=\frac{x^2y+y+x^2y^3}{x(1+x^2y^2-x^2)}\\\\y'=\frac{y(x^2+1+x^2y^2)}{x(-x^2+1+x^2y^2)}$

$2)\; \; x-3y+e^{y}-5=0\\\\x'-3y'+e^{y}\cdot y'-0=0\\\\y'\cdot (e^{y}-3)=-1\\\\y'=-\frac{1}{e^{y}-3}\\\\y'=\frac{1}{3-e^{y}}\\\\3)\; \; \left \{ {{x=ln\frac{t^2-1}{4}} \atop {y=sint}} \right.\\\\x'_{t}=\frac{4}{t^2-1}\cdot \frac{2t}{4}=\frac{2t}{t^2-1}\\\\y'_{t}=cost\\\\y'_{x}=\frac{y'_{t}}{x'_{t}}=\frac{(t^2-1)\cdot cost}{2t}$