Let $x_0$ and $x_1$ be two distinct nodes. Let $P(x)$ be a polynomial of degree 2 or less such that :
$P_0' (x_0 ) = f' (x_0 )$ , $P (x_1 ) = f (x_1 )$ , $P_0' (x_1 ) = f' (x_1 )$ , and $P (x) = f' (x_0 )Q_0 (x) + f (x_1 )Q_1 (x) + f' (x_1 )Q_2 (x)$
for some polynomials $Q_0 , Q_1 , Q_2$.
Write down the conditions that determine $Q_0 , Q_1 , Q_2$ (e.g. $Q_1 (x_1 ) = 1$) and then determine $Q_0 , Q_1 , Q_2$ .
My initial approach was to find interpolation of $P'(x)$ from first and third equation then get $P(x)$ by integrating $P'(x)$ determine the constant value from the second equation. But I am not sure if this is the correct approach.
Let's just follow the advice (the second paragraph you wrote: "Write down ..."). We have $$ P'(x) = f'(x_0)Q_0'(x) + f(x_1)Q_1'(x) + f'(x_1)Q_2'(x) $$ And hence \begin{align*} f'(x_0) &\stackrel != P'(x_0) = f'(x_0)Q_0'(x_0) + f(x_1)Q_1'(x_0) + f'(x_1)Q_2'(x_0)\\ f(x_1) &\stackrel != P(x_1) = f'(x_0)Q_0(x_1) + f(x_1)Q_1(x_1) + f'(x_1)Q_2(x_1)\\ f'(x_1) &\stackrel != P'(x_1) = f'(x_0)Q_0'(x_1) + f(x_1)Q_1'(x_1) + f'(x_1)Q_2'(x_1) \end{align*} Hence, we must have \begin{align*} Q_0'(x_0) &= 1 & Q_0(x_1) &= 0 & Q_0'(x_1) &= 0\\ Q_1'(x_0) &= 0 & Q_1(x_1) &= 1 & Q_1'(x_1) &= 0\\ Q_2'(x_0) &= 0 & Q_2(x_1) &= 0 & Q_2'(x_1) &= 1 \end{align*} Now, $Q_0'$ is a degree one polynomial, with $Q'_0(x_0) = 1$ and $Q'_0(x_1) = 0$, giving $$ Q'_0(x) = \frac{x-x_1}{x_0-x_1} $$ As you say, now for some constant $$ Q_0(x) = \frac 12\frac{(x-x_1)^2}{x_0-x_1} + c_0 $$ $Q_0(x_1) = 0$ gives $c_0 = 0$, hence $$ Q_0(x) = \frac 12\frac{(x-x_1)^2}{x_0-x_1} $$ $Q'_1$ has two zeros, but degree at most one, hence $Q'_1 = 0$. So, $Q_1$ is a constant, and $Q_1(x_1) = 1$, so $$ Q_1(x) = 1. $$ For $Q_2$ the same argument as for $Q_0$ but with the roles of $x_0$ and $x_1$ exchanged gives $$ Q_2(x) = \frac 12\frac{(x-x_0)^2}{x_1 - x_0}. $$