Diagonaliz the matrix

All I know is what I just got from Wiki after reading your question. Below is my attempt and you need to double check with what you learnt from class, school, etc.
Step 1 : we have to first find whether there are eigenvalues (λ) for A and if so, how many. λ will satisfy the following equation :
Ax= λx for one or some non-zero vector x.
It means Ax= λIx where I is the identity matrix
or (A-λI)(x)=0
Since x needs to be non-zero, det(A-λI)=o. That is, the determinant of the following matrix = 0 (Note : very diffcult to type a matrix here).
(2-λ) 4 0
0 (1-λ) 0 =0
3 3 (1-λ)

(2-λ)(-1-λ)(1-λ) =0
Therefore, λ=-1, 1 or 2.
Since A is a 3x3 matrix and there are three eigenvalues, A is diagonalizable.
Step 2 : compute the vectors (and then take the one with simplest form) for each of the eigenvalues.
When =1 x =[0,0,c] (vertically placed) where c is a constant. Take c=1 for easy computation and v1=[0,0,1] (vertically placed);
When =-1, x=[8c,-6c,-3c] where c is a constant. Take c=1 for easy computation and v2=[8,-6,-3](vertically placed);
When = 2, x=[c,0, -3c] where c is a constant. Take c=1 for easy computation and v3=[1,0,-3] (vertically placed).
So, the matrix P is constructed by putting these three vectors, as follows:
0 8 1
0 -6 0
1 -3 -3
And P^(-1) (your D?) (i.e. the inverse of P) can be computed out as follows:
(-3) (-9/2) 1
0 (-1/6) 0
1 (4/3) 0
Step : Checking
It is necessary to check PP^(-1)=P^(-1)P=I. I have checked it and leave it to you to re-check it as exercise. One is also required to compute P^(-1)AP to double check whether the product is a diagonal matrix. My computation shows that it is OK as follows:
1 0 0
0 (-1) 0
0 0 2
Hope the above is of assistance to you.