SymPy diff () function problem

I am working on implementing various numeric methods in python including reverse Euler. Thus, I decided to implement the fast Newton-Raphson method, which would give me the desired root for the variables of the equations that I would have to deal with with the calculator. Here he is:

def Newton(y0, f, tol=1e-10):
    y = Symbol('y')
    df = diff(f, y)
    while f.subs(y, y0) > tol:
        y0 -= f.subs(y, y0)/df.subs(y, y0)
    return y0

      

        
        
        
      

    

A code excerpt that includes a function call Newton()

:

def InverseEuler(f, y0, t0, tf, h):
    coords = [[t0, y0]]
    t, y = symbols('t y')

    while round(t0, 7) < round(tf, 7):
        aux = sympify(y0 - y + h*f.subs(t, t0 + h))
        print aux
        y0 = utils.Newton(y0, aux)
        t0 += h
        coords.append([t0, y0])
    return coords

      

        
        
        
      

    

Where f

is the "sympified" symbol. In my failed test case, it was:f = sympify('(y**2 + t)/(y - t)')

As you can see, I am printing aux

content to keep track of exactly where the diff function is not working. This was the first iteration for y (0) = 1 for h = 0,1, where aux: -y + 1 + 0.1*(y**2 + 0.1)/(y - 0.1)

. Going through aux

both f

in Newton()

and then running diff(f, y)

it gives me:

Traceback (most recent call last):
  File "C:\Users\Gabriel Vasconcelos\Documents\python\Metodos\metodos.py", line 97, in <module>
    GPHandler.replot(InverseEuler(f, 1, 0, 4, 0.1))
  File "C:\Users\Gabriel Vasconcelos\Documents\python\Metodos\metodos.py", line 31, in InverseEuler
    y0 = utils.Newton(y0, aux)
  File "C:\Users\Gabriel Vasconcelos\Documents\python\Metodos\utils.py", line 6, in Newton
    df = diff(f, y)
  File "C:\Python27\lib\site-packages\sympy\mpmath\calculus\differentiation.py", line 188, in diff
    values, norm, workprec = hsteps(ctx, f, x, n, prec, **options)
  File "C:\Python27\lib\site-packages\sympy\mpmath\calculus\differentiation.py", line 61, in hsteps
    values = [f(x+k*h) for k in steps]
TypeError: 'Add' object is not callable

      

        
        
        
      

    

Surprisingly, when I manually 'simpify' the same equation and 'diff' it, it works. Newton's method also works like a plume. However, I don't know what might be wrong.