$$y_{n+1} = y_n +hf(h,y_{n+1})$$
How do I use implicit Euler to solve this $y(x_1)$?
$$y' = \frac{-x}{y^2}$$ $$y(0) = 1$$ $$h=0.1 $$ $$x_1=0.1$$
I have got this far:
$$y_1 = y_0 + hf(h,y_1) = y_0-\frac{h^2}{y_1^2}$$
But I do not how to sepeatete the $y_1$ on the left side. Can somebody help me?
Newton's method is a possibility, as mentioned in the comments, but the equation to be solved is already in the form of a fixed point problem. You want to compute $y_{n+1}$, solution of $$ y_{n+1} = \underbrace{y_n + h f(t_{n+1}, y_{n+1})}_{g(y_{n+1})} $$
So, the fixed point method can start from $y_{n+1}^{(0)} = y_n$ and use the iteration $y_{n+1}^{(k+1)} = y_n + h f(t_{n+1},y_{n+1}^{(k)})$.