Consider the set of polynomials {${\phi_n}$}built using the Gramm-Schmidt method with respect to the inner product:
$$(f,g) = \int_{a}^{b} f(x)g(x)w(x)dx$$
with $w(x)>0$.
Prove that if $x_0$ is a zero of $\phi_n$, then:
(1) $$x_0 = \frac{\int_{a}^{b}x[\frac{\phi_n}{x-x_o}]^2 w(x)dx}{\int_{a}^{b}[\frac{\phi_n}{x-x_o}]^2 w(x)dx}(1)$$
From that,I've got to this:
$$\int_{a}^{b} \phi_n(x)\frac{ \phi_n(x)}{x-x_0}w(x)dx = 0$$ I suspect this is the solution but I don't know how to argue it. The second term inside the integral is of one degree less than $\phi_n$, by the properties of the polynomials, would that make it orthogonal to $\phi_n$?.
Also, the problem then says: find an analogue formula for $x_0^2$ pertaining to (1), which I don't understand, maybe someone gets it.
You are on the right track. All you need to say is that $\frac{\phi_n}{x-x_0}$ is a polynomial of degree $n-1$. Since $x_0$ is a root, you have $\phi_n(x)=(x-x_0)P_{n-1}(x)$. Now we can write $P_{n-1}$ in terms of the orthogonal basis, and we only need the polynomials with degree less or equal to $n-1$:$$P_{n-1}(x)=\sum_j^{n-1}c_j\phi_j(x)$$Here $c_j$ are just some constants. When you plug in this expression into the left hand side of your last formula, you get $$\sum_j^{n-1}c_j\int_a^b\phi_n(x)\phi_j(x)w(x)dx$$ Since the $\{\phi_n\}$ polynomials are orthogonal and $j\ne n$, the last integral is always $0$.
Now for the last question, you want to find a formula for $x_0^2$. Start from your last formula.
$$\int_{a}^{b} \phi_n(x)\frac{ \phi_n(x)}{x-x_0}w(x)dx = 0$$ Now multiply and divide by $(x-x_0)^2=x^2-2xx_0+x_0^2$:
$$\int_{a}^{b} \phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^3}(x^2-2xx_0+x_0^2)w(x)dx = 0$$ Expanding the parentheses, and moving the $x_0^2$ part to the other side, we get
$$x_0^2\int_a^b\phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^3}w(x)dx=\int_a^b(2xx_0-x^2)\phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^3}w(x)dx$$ Now use $2xx_0-x^2=x^2-2x(x-x_0)$, and you get $$x_0^2\int_a^b\phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^3}w(x)dx=\int_a^bx^2\phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^3}w(x)dx-2\int_a^bx\phi_n(x)\frac{ \phi_n(x)}{(x-x_0)^2}w(x)dx$$