A question about differential function If $f(x)=f'(x^{2})+2x$, then $f(1)=?$ and $f''(1)=?$
Sorry. I am going to check the original problem, and then i will update.

Consider the equation $f(x) = f'(x^2) + 2x$. It can be seen that $f(1) = 4$, $f'(1) = 2$ and $f^{(n+2)}(1) = 0$ for $n \geq 0$. The demonstration of how these values are obtained is as follows.

Since $f(x)$ is differentiable, which is evident from the differential equation, then consider the function in a series expansion given by
\begin{align}
f(x) = \sum_{n=0}^{\infty} a_{n} \, x^{n}
\end{align}
for which the equation given leads to
\begin{align}
a_{0} + a_{1} \, x + a_{2} \, x^2 + a_{3} \, x^3 + \cdots &= 2x + (a_{1} + 2 \, a_{2} \, x^2 + 3 \, a_{3} \, x^4 + 4 \, a_{4} \, x^6 + \cdots ) \\
&= a_{1} + 2 x + 2 \, a_{2} \, x^2 + 3 \, a_{3} \, x^4 + 4 \, a_{4} \, x^6 + \cdots .
\end{align}
Equating coefficients provides $a_{0} = a_{1}$, $a_{1} = 2$, $a_{n+ 2} = 0$ for $n \geq 0$. Hence 
\begin{align}
f(x) = 2 (x+1).
\end{align}
Now it is easy to obtain $f'(x) = 2$ and $f^{(n+2)}(x) = 0$ for $n \geq 0$. Letting $x=1$ provides the desired results. 
