I'm given a function in two variables $\;f(x,y)\;$ which has continuous partial derivatives of order $\;2\;$ in some open neighborhood $\;D\;$ of $\;(0,0)\;$ . We're also given that in $\;D\;$ :
i) $\;f_x(x,y)+xf_y(x,y)= f(x,y)\;$
ii) $\;f(t^2,t)=t\;$
We have to write the Taylor polynomial of order two around $\;(0,0)\;$ .
Now my work: since $\;f\;$ is continuous in $\;(0,0)\;$ we have from (ii) that
$$f(0,0)=\lim_{t\to 0}f(t^2,t)=\lim_{t\to 0}t=0$$
since the limit must exist no matter how we approach the origin.
We also have, by continuity of the partial derivatives, that taking $\;y=x\;$ in (i) we get
$$0=f(0,0)=\lim_{x\to 0}f(x,x)=\lim_{x\to 0}\left(f_x(x,x)+xf_y(x,x)\right)=f_x(0,0) +0\cdot f_y(0,0)=f_x(0,0)$$
Now, differentiating (i) wrt $\;x\;$ and then wrt $\;y\;$ , we get, using continuity at origin:
$$\begin{align*}I&\;\;f_{xx}(x,y)+f_y(x,y)+xf_{xy}(x,y)=f_x(x,y)\implies\\ \;&f_{xx}(0,0)+f_y(0,0)+0\cdot f_{xy}(0,0)=f_x(0,0)=0\implies\\ \;&f_{x^2}(0,0)=-f_y(0,0)\\{}\\ II&\;\;f_{xy}(x,y)+xf_{y^2}(x,y)=f_y(x,y)\implies\\ \;&f_{xy}(0,0)=f_y(0,0)\end{align*}$$
The Taylor polynomial wanted is then (in $\;D\;$ )
$$f(x,y)=f(0,0)+f_x(0,0)x+f_y(0,0)y+\frac{f_{x^2}(0,0)x^2}2+\frac{f_{y^2}(0,0)y^2}2+f_{xy}(0,0)xy=$$
$$=f_y(0,0)\left(y-\frac12x^2+xy\right)+\frac12f_{y^2}(0,0)y^2$$
...and here I got stuck .
Added self work: I've been thinking if we can do the following:
$$\begin{cases}x(t):=t^2\\y(t):=t\end{cases}\;\;\stackrel{(ii)}\implies\;t=f(t^2,t)=f(x(t),y(t))$$
So now we can apply chain rule in the above:
$$1=\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}=f_x\cdot 2t+f_y\cdot 1$$
and if we now take the limit and use continuity and other result above, we get $\;f_y(0,0)=1\;$
Is this correct? If it is I'm left only with the task of finding $\;f_{y^2}(0,0)\;$ , which I don't know yet.
Added: Any idea about the last self work will be greatly appreciated.