Найти y 3 порядка функции y=f(x), заданой неявно:/ x^2+2xy+y^2-4x+2y-2=0, в точке А(1,1)

[latex]x^2+2xy+y^2-4x+2y-2=0\; ,\; \; A(1.,1)\\\\(x^2+2xy+y^2-4x+2y)'=2'\\\\2x+(2x)'y+(2x)y'+2yy'-4+2y'=0\\\\2x+2y+2xy'+2yy'+2y'-4=0\\\\2y'(x+y+1)=2(2-x-y)\\\\ y'=\frac{2-x-y}{x+y+1} \\\\y''= \frac{(-1-y')(x+y+1)-(2-x-y)(1+y')}{(x+y+1)^2} =\\\\= \frac{-x-y-1-y'(x+y+1)-2+x+y-y'(2-x-y)}{(x+y+1)^2} = \frac{-y'(x+y+1+2-x-y)-3}{(x+y+1)^2} =\\\\= \frac{-3-3y'}{(x+y+1)^2}= \frac{-3-\frac{-6+3x+3y}{x+y+1}}{(x+y+1)^2} = \frac{-9}{(x+y+1)^3} [/latex] [latex]y'''= \frac{9\cdot 3(x+y+1)^2\cdot (1+y')}{(x+y+1)^6} = \frac{27(1+y')}{(x+y+1)^4} =\\\\= \frac{27(1+\frac{2-x-y}{x+y+1})}{(x+y+1)^4}= \frac{27(x+y+1+2-x-y)}{(x+y+1)^5} = \frac{81}{(x+y+1)^5} \\\\y'''(1,1)=\frac{81}{3^5}=\frac{3^4}{3^5}=\frac{1}{3}[/latex]