Substitution in implicit function

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I am not completely sure about your question, but i guess that what you call "procedure" is actually a Python function, which you can define using the def statement (check for Python introductions, there are very good tutorials online):

var("w1, w2, a, p")
EU(w1, w2, a, p)= p*w1^a+ (1-p)*w2^a

def my_procedure(V): 
    var("dw1, dw2")
    V_w1 = diff(V, w1)
    V_w2 = diff(V, w2)
    # Differential
    dV = V_w1 * dw1 + V_w2 * dw2 
    show("dV ="+latex(dV))
    # Dérivée du premier ordre
    sol = solve(dV == 0, dw2)
    show(sol[0]/dw1)

Then, calling:

my_procedure(EU)

leads to

$dV = \left( w_{1}, w_{2}, a, p \right) \ {\mapsto} \ a \mathit{dw}_{1} p w_{1}^{a - 1} - a \mathit{dw}_{2} {\left(p - 1\right)} w_{2}^{a - 1}$

$\frac{\mathit{dw}_{2}}{\mathit{dw}_{1}} = \frac{p w_{1}^{a - 1} w_{2}^{-a + 1}}{p - 1}$

and

V=function('V')(w1, w2)
my_procedure(V)

leads to

$dV = \mathit{dw}_{1} \frac{\partial}{\partial w_{1}}V\left(w_{1}, w_{2}\right) + \mathit{dw}_{2} \frac{\partial}{\partial w_{2}}V\left(w_{1}, w_{2}\right)$

$\frac{\mathit{dw}_{2}}{\mathit{dw}_{1}} = -\frac{\frac{\partial}{\partial w_{1}}V\left(w_{1}, w_{2}\right)}{\frac{\partial}{\partial w_{2}}V\left(w_{1}, w_{2}\right)}$