How can I find the cost function $c(w,p)$ given that the production is
$$ f(x)=(x_1^p + x_2^p)^{1/p} \ \ for\ \ 0<p <1 $$
I tried to solve it and found that $$TC(y) = \left\{ \begin{array}{ll} w_1y & \quad w_1 < w_2 \\ wy & \quad w_1=w_2 \\ w_2y & \quad w_2 <w_1 \end{array} \right. $$
Can you say me if I'm on the right way?
If you are interested in the case where $\rho \geq 1$ then look at the post CES $\ \ \rho \geq 1$. For the standard case where $0 < \rho < 1$ you should get a result like this
$$C(w_1,w_2,y) = \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y.$$
To see this you should start by setting up the cost minimization problem
$$\min_{x_1,x_2} \ \ w_1x_1 + w_2x_2 \\[8pt] s.t. \ \ (x_1^\rho + x_2^\rho)^{1/\rho} \geq y$$
for this problem the Lagrangian function is
$$\mathcal L(x_1,x_2,\lambda) = w_1x_1 + w_2x_2 - \lambda((x_1^\rho + x_2^\rho)^{1/\rho} -y).$$
From the first order conditions of the Lagrangian you can show the constraint is binding in optimum $(x_1^\rho + x_2^\rho)^{1/\rho} = y$ and get MRS equal to relative prices
$$(1) \ \ \frac{w_1}{w_2} = \frac{x_1^{\rho - 1}}{x_2^{\rho - 1}},$$ given this information you should be able to solve for $x_1$ and $x_2$ as a function of the parameters of the problem which in this case is $\rho,y,w_1,w_2$.
Try to get $(x_1^\rho + x_2^\rho)^{1/\rho}$ to appear in the MRS equal to relative prices. So manipulate (1) to get
$$w_1^{\frac{\rho}{\rho -1}}x_2^\rho = w_2^{\frac{\rho}{\rho -1}}x_1^\rho,$$ then add $w_2^{\frac{\rho}{\rho-1}}x_2^\rho$ to both sides of equation
$$w_1^{\frac{\rho}{\rho -1}}x_2^\rho +w_2^{\frac{\rho}{\rho -1}}x_2^\rho = w_2^{\frac{\rho}{\rho -1}}x_1^\rho + w_2^{\frac{\rho}{\rho -1}}x_2^\rho,$$
isolate factors on both sides and exponentiate with exponent $1/\rho$ to get
$$\left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}x_2 = w_2^{\frac{1}{\rho -1}}(x_2^\rho + x_1^\rho)^{1/\rho} = w_2^{\frac{1}{\rho -1}} y ,$$
from here you can solve for conditional demand $x_2^\star(w_1,w_2,y)$. However, it is easier to oberserve that the factor $\left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$ do not change when interchanging indexes - it is symmetric. So define $a := \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{1/\rho}$ and conclude that
$$ax_1 = w_1^{\frac{1}{\rho -1}} y \\[8pt] ax_2 = w_2^{\frac{1}{\rho -1}} y,$$ multiply first equation with $w_1$ and second with $w_2$ and add them to get
$$a(w_1x_1 + w_2x_2) = (w_1^{\frac{\rho}{\rho -1}}+w_2^{\frac{\rho}{\rho -1}})y = a^\rho y$$
solve for $(w_1x_1 + w_2x_2)$ which are the costs to get the result that
$$C(w_1,w_2,y) = a^{\rho -1} y = \left(w_1^{\frac{\rho}{\rho -1}} +w_2^{\frac{\rho}{\rho -1}}\right)^{\frac{\rho - 1}{\rho}} y$$