Intersection of the blowing up and the exceptional curve for a concrete example Let $f(x,y)=x^2y+xy^2 - x^4-y^4=0$ be an affine curve.
Consider its blowing up at the origin, namely $x^2y+xy^2 = x^4+y^4$  and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
If $t \neq 0$ then we can set $y=xu$ and substituting this  to the equation of curve, we obatin $x^3(2u-x(1+u^4))=0$, which implies the intersection of the blowing up and the exceptional curve is the point $u=0,x=0$. Because $f(x,y)=f(y,x)$, we  get the same  result for the case $ u \neq 0$ with $x=ty$, namely, we obtain   $t=0,y=0$. Thus intersection of the blowing up and the exceptional curve is constructed by these two points. Is it correct?

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I was wrong.
Case I. 
$u=1$ and $x^2y+xy^2 = x^4+y^4$  and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
Then, $x^3\{u+u^2-x(1+u^4)\}=0$ and thus the intersection of exceptional curve $\{x=0\}$ and the blowing up is given by the equation $  u+u^2-0\times(1+u^4)=0$. Thus $u=0$ and $u=-1$ are the intersection.
Case II.
$t=1$ and $x^2y+xy^2 = x^4+y^4$  and $xu=ty$ in $\mathbb{A}^2 \times \mathbb{P}^1$.
Then, $y^3\{t+t^2-y(1+t^4)\}=0$ and thus the intersection of exceptional curve $\{y=0\}$ and the blowing up is given by the equation $  t+t^2-0\times(1+t^4)=0$. Thus $t=0$ and $t=-1$ are the intersection.
Because $(t:u)=(1:-1)=(-1:1)$, $t=-1$ in case II and  $u=-1$ in case I are same points,  thus the intersection of the blowing up and the exceptional curve consist three distinct points.

As Lazzaro Campoetti says, you've made a mistake in your calculations. After substituting $y = xz$ and simplifying, I find the equation
$$
0 = x^3(z(1+z) - x(1+z^4)) \, .
$$
The factor $x^3$ corresponds to the exceptional locus. However, it is not a point, but a line: since $y = xz$, then $x = 0$ implies $y = 0$ and leaves $z$ unconstrained. Thus the exceptional locus is the $z$-axis, i.e., the vertical line $x = 0, y = 0$.
The other factor $z(1+z) - x(1+z^4)$ is the strict transform of the curve. Since $y = xz$, setting $z(1+z) - x(1+z^4) = 0$ we obtain the parametric equation
$$
(x,y,z) = \left(\frac{z(1+z)}{1+z^4}, \frac{z^2(1+z)}{1+z^4}, z \right) \, .
$$
To find the intersection with the exceptional locus, we must find the points where $x = 0, y = 0$. Since $y = xz$, then it suffices to solve $0 = x = \frac{z(1+z)}{1+z^4}$. Thus we find that the strict transform intersects the exceptional locus when $z = 0$ and $z = -1$. However, there is also one more intersection at infinity, since
$$
x = \frac{z(1+z)}{1+z^4} \to 0 \quad \text{as $z\to \infty$} \, .
$$
(You could also show this by setting $z = Z_1/Z_0$, clearing denominators, and then setting $Z_0 = 0$.) You can see these features in the plot in this SageMath cell.
Moreover, based on considerations of the tangent directions at the singular point $(0,0)$ of the original curve, you should expect there to be $3$ intersections. The coordinate $z = y/x$ keeps track of the tangent slopes, and as I mentioned in this comment, the tangent lines at the origin and $x = 0$, $y = 0$, and $y = -x$, which have slopes $\infty$, $0$, and $-1$.
