I'm stuck at the first part. I think the image of this morphism should be described by three polynomial equations. I think I found one, which is xw-yz=0 where (x, y, z, w) is the coordinate of K^4. However, I fail to find other two equations. Could anyone help me? Also could anyone show me why X is not isomorphic to any curve in K^3?
You want to find equations in $x,y,z,w$ which express the fact that a point satisfying them is of the form $(t^4,t^5,t^6,t^7)$ for some $t$.
There are two approaches to this: the first is to observe that, when $x\neq 0,$ obviously $t = y/x$, and so we want the equations to be $x = (y/x)^4,$ $y = (y/x)^5$, $z = (y/x)^6$, and $w = (y/x)^7,$ when $x$ is non-zero.
But these equations have $x$ in the denominator, and so to get equations on all of $\mathbb A^4$, we have to clear denominators. But, as we will see, this can lead to spurious solutions with $x =0$.
If we clear denominators in the above equations, we get the equations $x^5 = y^4, x^6 z = y^6, \text{ and } x^7 w = y^7.$ (The second of the above equations reduces to the first one if we clear denominators and divide through by $y$.) If we take into account the first equation, we can simplify the second to $xz = y^2,$ and the third simplifies to $x^2 w = y^3 = y (y^2) = xyz,$ which we can further simplify to $xw = yz$ by dividing through by $x$.
So far, we have obtained the equations $x^5 = y^4,$ $xz = y^2$, and $xw = yz.$ But these equations still don't cut out the image you want; they also have the whole $2$-plane $(0,0,z,w)$ in their zero-locus.
We could manipulate the equations we have more, and try to do more divisions by $x$ and $y$ to eliminate spurious solutions, but at this point it may be easier to note the alternative approach: just look for combinations of the exponents $4,5,6,$ and $7$ that are equal to one another.
E.g. as noted in comments it also satisfies $x^3 = z^2$ and $xz^4 = w^4$. Actually, there is also a lower degree equation involving $x,z,$ and $w$, namely $x^2 z = w^2$. (As I will show, we can recover $xz^4 = w^4$ from this and the other equations we already have.) Note that the first of these, when combined with $xz = y^2$, gives back $x^5 = y^4$ (our starting point), so now we have the equations $$ x^3 = z^2, xz = y^2, x^2z = w^2, xw = yz.$$ (To recover $xz^4 = w^4$, note that starting from $xz = y^2$, we get $xz^4 = y^2 z^3 = x^2 w^2 z = w^4.$)
Certainly, from these equations, we see that if $x = 0$ then also $y = 0 z = w = 0$. If $x\neq 0,$ then defining $t = y/x$, you can work backwards through the above computations to check that $y = t^5, etc. Thus these equations exactly cut out the image of the original morphism, and so show that it is an algebraic set.
That this algebraic set is irreducible and one-dimensional follows from easy topological arguments, using the fact that the domain of the morphism is one-dimensional and irreducible. To see that it can't be embedded into affine $3$-space, one approach is to compute the Zariski tangent space at the origin and show that it is four dimensional. For this, you really need to compute the full ideal of functions vanishing on the algebraic curve.
What I mean is: if we denote by $V$ the image of the morphism, and by $I$ the ideal $(x^3 - z^2, xz - y^2, x^2 z - w^2, xw - yz),$ then we have shown that $V = V(I)$, but we haven't shown that $I = I(V)$. But to compute the Zariski tangent space to the curve at the origin, you need to know the local ring of the curve at the origin, and for this you need to know $I(V)$. In fact there are some additional equations in $I(V)$; if I'm not mistaken, we have $I(V) = (x^3 - z^2, xz - y^2, x^2 z - w^2, xw - yz, yw - z^2, z w - yx^2).$ I'll leave this for you to check.
From this you can compute that the Zariski tangent space of $V$ at the origin really is $4$-dimensional.