<s>[INST] Derivation and tangent problem set Problem 1
On the curve $y=\frac{1}{1+x^2}$ find a point in which tangent line is parallel to the horizontal axis.

My idea:

Let's find $y'$. 
$$y'=\frac{-2x}{(1+x^2)^2}$$
If we want a tangent line to be parallel to x-axis it must look like this $t...y=kx+l$ such that $k=0$.
Let's find for which $x$ we have that situation.
$$y'=\frac{-2x}{(1+x^2)^2}=0$$
$$x=0\to y=1$$
$$(x,y)=(0,1)$$
Problem 2
Determine the angle of the curve
$y^2 - 3x^2 + x + 1 = 0$ and 
$xy^2 - 1 = 0$
intersect in the first quadrant
My work:
$y^2 - 3x^2 + x + 1 =xy^2 - 1$
One solution for $x>0$ and $y>0$ is $x=y=1$.
Let's now find derivative of easch function at that point.
$$2xyy'=-y^2$$
$$y'=\frac{-y^2}{2xy}=\frac{-1}{2}$$
$$2yy'=6x-1$$
$$y'=\frac{6x-1}{2y}=\frac{5}{2}$$
And the angle is $\alpha =arctg|\frac{k_1-k_2}{1+k_1*k_2}|$ where $k_1=\frac{-1}{2}$ and $k_2=\frac{5}{2}$

Problem 3
Find the angle of intersection of the tangents to the curve
$3x^2 - 6x + 2y^2 - 3 = 0$
from the point $T( 1,3 )$?
My work:
First (1,3) isn't on the curve.
We will need that tangent equation:
$$y-y_0=y'(x_0)(x-x_0)$$
From the task we know that $x=1$ and $y=3$.
Let's try to find $x_0$ and $y_0$. 
Using implict differentation we will find $y'$.
$$6x-6+4yy'=0$$
$$3x-3+2yy'=0$$
$$2yy'=-3x+3$$
$$y'=\frac{-3x+3}{2y}$$
Plugin that in equation:
$$3-y_0=\frac{-3x_0+3}{2y_0}(1-x_0)$$
$$3-y_0=\frac{-3x_0+3}{2y_0}+\frac{-3x_0^2-3x_0}{2y_0}$$
$$3-y_0=\frac{-3x_0^2+3-6x_0}{2y_0}$$
$$6y_0-2y_0^2=-3x_0^2+3-6x_0$$
But I can't get something nice...Is my problem 1 and 2 ok and please help me with 3
 [/INST] Problem  1 is OK.
Problem  2 ;
First try to find intersection point $(x,y)$
$$ \frac1x - 3 x^2 + x +1  =0 $$
Factorize it 
$$ -(x-1) (3 x^2  2 x +1) =0 $$
At $x=1, y= \dfrac{1}{\sqrt x } =1.\; i.e., (1,1) $ is the point of intersection. Other roots are neither in first quadrant nor real.
Now for the slope: $ y^2 = \frac 1x \rightarrow y'= \dfrac{-1}{ x^2 y} \rightarrow -1. $
Problem 3:
not answered.
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